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<!DOCTYPE html PUBLIC "-//IETF//DTD HTML 2.0//EN">
<HTML>
<HEAD>
<TITLE>Stanford ACM-ICPC Team Notebook</TITLE>
</HEAD>
<BODY BGCOLOR="#FFFFFF" TEXT="#000000" LINK="#1F00FF" ALINK="#FF0000" VLINK="#9900DD">
<A NAME="top">
<CENTER><H1><U>Stanford ACM-ICPC Team Notebook</U></H1></CENTER>
<H1>Table of Contents</H1>
<H2>Combinatorial optimization</H2>
<OL START=1>
<LI><A HREF="#file1">Sparse max-flow</A></LI>
<LI><A HREF="#file2">Min-cost max-flow</A></LI>
<LI><A HREF="#file3">Push-relabel max-flow</A></LI>
<LI><A HREF="#file4">Min-cost matching</A></LI>
<LI><A HREF="#file5">Max bipartite matchine</A></LI>
<LI><A HREF="#file6">Global min-cut</A></LI>
<LI><A HREF="#file7">Graph cut inference</A></LI>
</OL>
<H2>Geometry</H2>
<OL START=8>
<LI><A HREF="#file8">Convex hull</A></LI>
<LI><A HREF="#file9">Miscellaneous geometry</A></LI>
<LI><A HREF="#file10">Java geometry</A></LI>
<LI><A HREF="#file11">3D geometry</A></LI>
<LI><A HREF="#file12">Slow Delaunay triangulation</A></LI>
</OL>
<H2>Numerical algorithms</H2>
<OL START=13>
<LI><A HREF="#file13">Number theory (modular, Chinese remainder, linear Diophantine)</A></LI>
<LI><A HREF="#file14">Systems of linear equations, matrix inverse, determinant</A></LI>
<LI><A HREF="#file15">Reduced row echelon form, matrix rank</A></LI>
<LI><A HREF="#file16">Fast Fourier transform</A></LI>
<LI><A HREF="#file17">Simplex algorithm</A></LI>
</OL>
<H2>Graph algorithms</H2>
<OL START=18>
<LI><A HREF="#file18">Fast Dijkstra's algorithm</A></LI>
<LI><A HREF="#file19">Strongly connected components</A></LI>
<LI><A HREF="#file20">Eulerian path</A></LI>
</OL>
<H2>Data structures</H2>
<OL START=21>
<LI><A HREF="#file21">Suffix array</A></LI>
<LI><A HREF="#file22">Binary Indexed Tree</A></LI>
<LI><A HREF="#file23">Union-find set</A></LI>
<LI><A HREF="#file24">KD-tree</A></LI>
<LI><A HREF="#file25">Splay tree</A></LI>
<LI><A HREF="#file26">Lazy segment tree</A></LI>
<LI><A HREF="#file27">Lowest common ancestor</A></LI>
</OL>
<HR>
<A NAME="file1">
<H1>code/Dinic.cc 1/27</H1>
[<A HREF="#top">top</A>][prev][<A HREF="#file2">next</A>]
<PRE>
<I><FONT COLOR="#B22222">// Adjacency list implementation of Dinic's blocking flow algorithm.
</FONT></I><I><FONT COLOR="#B22222">// This is very fast in practice, and only loses to push-relabel flow.
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// Running time:
</FONT></I><I><FONT COLOR="#B22222">// O(|V|^2 |E|)
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// INPUT:
</FONT></I><I><FONT COLOR="#B22222">// - graph, constructed using AddEdge()
</FONT></I><I><FONT COLOR="#B22222">// - source and sink
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// OUTPUT:
</FONT></I><I><FONT COLOR="#B22222">// - maximum flow value
</FONT></I><I><FONT COLOR="#B22222">// - To obtain actual flow values, look at edges with capacity > 0
</FONT></I><I><FONT COLOR="#B22222">// (zero capacity edges are residual edges).
</FONT></I>
#<B><FONT COLOR="#5F9EA0">include<cstdio>
</FONT></B>
#<B><FONT COLOR="#5F9EA0">include<vector>
</FONT></B>
#<B><FONT COLOR="#5F9EA0">include<queue>
</FONT></B>
using namespace std;
<B><FONT COLOR="#228B22">typedef</FONT></B> <B><FONT COLOR="#228B22">long</FONT></B> <B><FONT COLOR="#228B22">long</FONT></B> LL;
<B><FONT COLOR="#228B22">struct</FONT></B> Edge {
<B><FONT COLOR="#228B22">int</FONT></B> u, v;
LL cap, flow;
Edge() {}
Edge(<B><FONT COLOR="#228B22">int</FONT></B> u, <B><FONT COLOR="#228B22">int</FONT></B> v, LL cap): u(u), v(v), cap(cap), flow(0) {}
};
<B><FONT COLOR="#228B22">struct</FONT></B> Dinic {
<B><FONT COLOR="#228B22">int</FONT></B> N;
vector<Edge> E;
vector<vector<<B><FONT COLOR="#228B22">int</FONT></B>>> g;
vector<<B><FONT COLOR="#228B22">int</FONT></B>> d, pt;
Dinic(<B><FONT COLOR="#228B22">int</FONT></B> N): N(N), E(0), g(N), d(N), pt(N) {}
<B><FONT COLOR="#228B22">void</FONT></B> AddEdge(<B><FONT COLOR="#228B22">int</FONT></B> u, <B><FONT COLOR="#228B22">int</FONT></B> v, LL cap) {
<B><FONT COLOR="#A020F0">if</FONT></B> (u != v) {
E.emplace_back(Edge(u, v, cap));
g[u].emplace_back(E.size() - 1);
E.emplace_back(Edge(v, u, 0));
g[v].emplace_back(E.size() - 1);
}
}
<B><FONT COLOR="#228B22">bool</FONT></B> BFS(<B><FONT COLOR="#228B22">int</FONT></B> S, <B><FONT COLOR="#228B22">int</FONT></B> T) {
queue<<B><FONT COLOR="#228B22">int</FONT></B>> q({S});
fill(d.begin(), d.end(), N + 1);
d[S] = 0;
<B><FONT COLOR="#A020F0">while</FONT></B>(!q.empty()) {
<B><FONT COLOR="#228B22">int</FONT></B> u = q.front(); q.pop();
<B><FONT COLOR="#A020F0">if</FONT></B> (u == T) <B><FONT COLOR="#A020F0">break</FONT></B>;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> k: g[u]) {
Edge &e = E[k];
<B><FONT COLOR="#A020F0">if</FONT></B> (e.flow < e.cap && d[e.v] > d[e.u] + 1) {
d[e.v] = d[e.u] + 1;
q.emplace(e.v);
}
}
}
<B><FONT COLOR="#A020F0">return</FONT></B> d[T] != N + 1;
}
LL DFS(<B><FONT COLOR="#228B22">int</FONT></B> u, <B><FONT COLOR="#228B22">int</FONT></B> T, LL flow = -1) {
<B><FONT COLOR="#A020F0">if</FONT></B> (u == T || flow == 0) <B><FONT COLOR="#A020F0">return</FONT></B> flow;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> &i = pt[u]; i < g[u].size(); ++i) {
Edge &e = E[g[u][i]];
Edge &oe = E[g[u][i]^1];
<B><FONT COLOR="#A020F0">if</FONT></B> (d[e.v] == d[e.u] + 1) {
LL amt = e.cap - e.flow;
<B><FONT COLOR="#A020F0">if</FONT></B> (flow != -1 && amt > flow) amt = flow;
<B><FONT COLOR="#A020F0">if</FONT></B> (LL pushed = DFS(e.v, T, amt)) {
e.flow += pushed;
oe.flow -= pushed;
<B><FONT COLOR="#A020F0">return</FONT></B> pushed;
}
}
}
<B><FONT COLOR="#A020F0">return</FONT></B> 0;
}
LL MaxFlow(<B><FONT COLOR="#228B22">int</FONT></B> S, <B><FONT COLOR="#228B22">int</FONT></B> T) {
LL total = 0;
<B><FONT COLOR="#A020F0">while</FONT></B> (BFS(S, T)) {
fill(pt.begin(), pt.end(), 0);
<B><FONT COLOR="#A020F0">while</FONT></B> (LL flow = DFS(S, T))
total += flow;
}
<B><FONT COLOR="#A020F0">return</FONT></B> total;
}
};
<I><FONT COLOR="#B22222">// BEGIN CUT
</FONT></I><I><FONT COLOR="#B22222">// The following code solves SPOJ problem #4110: Fast Maximum Flow (FASTFLOW)
</FONT></I>
<B><FONT COLOR="#228B22">int</FONT></B> <B><FONT COLOR="#0000FF">main</FONT></B>()
{
<B><FONT COLOR="#228B22">int</FONT></B> N, E;
scanf(<B><FONT COLOR="#BC8F8F">"%d%d"</FONT></B>, &N, &E);
Dinic dinic(N);
<B><FONT COLOR="#A020F0">for</FONT></B>(<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < E; i++)
{
<B><FONT COLOR="#228B22">int</FONT></B> u, v;
LL cap;
scanf(<B><FONT COLOR="#BC8F8F">"%d%d%lld"</FONT></B>, &u, &v, &cap);
dinic.AddEdge(u - 1, v - 1, cap);
dinic.AddEdge(v - 1, u - 1, cap);
}
printf(<B><FONT COLOR="#BC8F8F">"%lld\n"</FONT></B>, dinic.MaxFlow(0, N - 1));
<B><FONT COLOR="#A020F0">return</FONT></B> 0;
}
<I><FONT COLOR="#B22222">// END CUT
</FONT></I></PRE>
<HR>
<A NAME="file2">
<H1>code/MinCostMaxFlow.cc 2/27</H1>
[<A HREF="#top">top</A>][<A HREF="#file1">prev</A>][<A HREF="#file3">next</A>]
<PRE>
<I><FONT COLOR="#B22222">// Implementation of min cost max flow algorithm using adjacency
</FONT></I><I><FONT COLOR="#B22222">// matrix (Edmonds and Karp 1972). This implementation keeps track of
</FONT></I><I><FONT COLOR="#B22222">// forward and reverse edges separately (so you can set cap[i][j] !=
</FONT></I><I><FONT COLOR="#B22222">// cap[j][i]). For a regular max flow, set all edge costs to 0.
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// Running time, O(|V|^2) cost per augmentation
</FONT></I><I><FONT COLOR="#B22222">// max flow: O(|V|^3) augmentations
</FONT></I><I><FONT COLOR="#B22222">// min cost max flow: O(|V|^4 * MAX_EDGE_COST) augmentations
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// INPUT:
</FONT></I><I><FONT COLOR="#B22222">// - graph, constructed using AddEdge()
</FONT></I><I><FONT COLOR="#B22222">// - source
</FONT></I><I><FONT COLOR="#B22222">// - sink
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// OUTPUT:
</FONT></I><I><FONT COLOR="#B22222">// - (maximum flow value, minimum cost value)
</FONT></I><I><FONT COLOR="#B22222">// - To obtain the actual flow, look at positive values only.
</FONT></I>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><cmath></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><vector></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><iostream></FONT></B>
using namespace std;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<<B><FONT COLOR="#228B22">int</FONT></B>> VI;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<VI> VVI;
<B><FONT COLOR="#228B22">typedef</FONT></B> <B><FONT COLOR="#228B22">long</FONT></B> <B><FONT COLOR="#228B22">long</FONT></B> L;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<L> VL;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<VL> VVL;
<B><FONT COLOR="#228B22">typedef</FONT></B> pair<<B><FONT COLOR="#228B22">int</FONT></B>, <B><FONT COLOR="#228B22">int</FONT></B>> PII;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<PII> VPII;
<B><FONT COLOR="#228B22">const</FONT></B> L INF = numeric_limits<L>::max() / 4;
<B><FONT COLOR="#228B22">struct</FONT></B> MinCostMaxFlow {
<B><FONT COLOR="#228B22">int</FONT></B> N;
VVL cap, flow, cost;
VI found;
VL dist, pi, width;
VPII dad;
MinCostMaxFlow(<B><FONT COLOR="#228B22">int</FONT></B> N) :
N(N), cap(N, VL(N)), flow(N, VL(N)), cost(N, VL(N)),
found(N), dist(N), pi(N), width(N), dad(N) {}
<B><FONT COLOR="#228B22">void</FONT></B> AddEdge(<B><FONT COLOR="#228B22">int</FONT></B> from, <B><FONT COLOR="#228B22">int</FONT></B> to, L cap, L cost) {
<B><FONT COLOR="#A020F0">this</FONT></B>->cap[from][to] = cap;
<B><FONT COLOR="#A020F0">this</FONT></B>->cost[from][to] = cost;
}
<B><FONT COLOR="#228B22">void</FONT></B> Relax(<B><FONT COLOR="#228B22">int</FONT></B> s, <B><FONT COLOR="#228B22">int</FONT></B> k, L cap, L cost, <B><FONT COLOR="#228B22">int</FONT></B> dir) {
L val = dist[s] + pi[s] - pi[k] + cost;
<B><FONT COLOR="#A020F0">if</FONT></B> (cap && val < dist[k]) {
dist[k] = val;
dad[k] = make_pair(s, dir);
width[k] = min(cap, width[s]);
}
}
L Dijkstra(<B><FONT COLOR="#228B22">int</FONT></B> s, <B><FONT COLOR="#228B22">int</FONT></B> t) {
fill(found.begin(), found.end(), false);
fill(dist.begin(), dist.end(), INF);
fill(width.begin(), width.end(), 0);
dist[s] = 0;
width[s] = INF;
<B><FONT COLOR="#A020F0">while</FONT></B> (s != -1) {
<B><FONT COLOR="#228B22">int</FONT></B> best = -1;
found[s] = true;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> k = 0; k < N; k++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (found[k]) <B><FONT COLOR="#A020F0">continue</FONT></B>;
Relax(s, k, cap[s][k] - flow[s][k], cost[s][k], 1);
Relax(s, k, flow[k][s], -cost[k][s], -1);
<B><FONT COLOR="#A020F0">if</FONT></B> (best == -1 || dist[k] < dist[best]) best = k;
}
s = best;
}
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> k = 0; k < N; k++)
pi[k] = min(pi[k] + dist[k], INF);
<B><FONT COLOR="#A020F0">return</FONT></B> width[t];
}
pair<L, L> GetMaxFlow(<B><FONT COLOR="#228B22">int</FONT></B> s, <B><FONT COLOR="#228B22">int</FONT></B> t) {
L totflow = 0, totcost = 0;
<B><FONT COLOR="#A020F0">while</FONT></B> (L amt = Dijkstra(s, t)) {
totflow += amt;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> x = t; x != s; x = dad[x].first) {
<B><FONT COLOR="#A020F0">if</FONT></B> (dad[x].second == 1) {
flow[dad[x].first][x] += amt;
totcost += amt * cost[dad[x].first][x];
} <B><FONT COLOR="#A020F0">else</FONT></B> {
flow[x][dad[x].first] -= amt;
totcost -= amt * cost[x][dad[x].first];
}
}
}
<B><FONT COLOR="#A020F0">return</FONT></B> make_pair(totflow, totcost);
}
};
<I><FONT COLOR="#B22222">// BEGIN CUT
</FONT></I><I><FONT COLOR="#B22222">// The following code solves UVA problem #10594: Data Flow
</FONT></I>
<B><FONT COLOR="#228B22">int</FONT></B> <B><FONT COLOR="#0000FF">main</FONT></B>() {
<B><FONT COLOR="#228B22">int</FONT></B> N, M;
<B><FONT COLOR="#A020F0">while</FONT></B> (scanf(<B><FONT COLOR="#BC8F8F">"%d%d"</FONT></B>, &N, &M) == 2) {
VVL v(M, VL(3));
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < M; i++)
scanf(<B><FONT COLOR="#BC8F8F">"%Ld%Ld%Ld"</FONT></B>, &v[i][0], &v[i][1], &v[i][2]);
L D, K;
scanf(<B><FONT COLOR="#BC8F8F">"%Ld%Ld"</FONT></B>, &D, &K);
MinCostMaxFlow mcmf(N+1);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < M; i++) {
mcmf.AddEdge(<B><FONT COLOR="#228B22">int</FONT></B>(v[i][0]), <B><FONT COLOR="#228B22">int</FONT></B>(v[i][1]), K, v[i][2]);
mcmf.AddEdge(<B><FONT COLOR="#228B22">int</FONT></B>(v[i][1]), <B><FONT COLOR="#228B22">int</FONT></B>(v[i][0]), K, v[i][2]);
}
mcmf.AddEdge(0, 1, D, 0);
pair<L, L> res = mcmf.GetMaxFlow(0, N);
<B><FONT COLOR="#A020F0">if</FONT></B> (res.first == D) {
printf(<B><FONT COLOR="#BC8F8F">"%Ld\n"</FONT></B>, res.second);
} <B><FONT COLOR="#A020F0">else</FONT></B> {
printf(<B><FONT COLOR="#BC8F8F">"Impossible.\n"</FONT></B>);
}
}
<B><FONT COLOR="#A020F0">return</FONT></B> 0;
}
<I><FONT COLOR="#B22222">// END CUT
</FONT></I></PRE>
<HR>
<A NAME="file3">
<H1>code/PushRelabel.cc 3/27</H1>
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<PRE>
<I><FONT COLOR="#B22222">// Adjacency list implementation of FIFO push relabel maximum flow
</FONT></I><I><FONT COLOR="#B22222">// with the gap relabeling heuristic. This implementation is
</FONT></I><I><FONT COLOR="#B22222">// significantly faster than straight Ford-Fulkerson. It solves
</FONT></I><I><FONT COLOR="#B22222">// random problems with 10000 vertices and 1000000 edges in a few
</FONT></I><I><FONT COLOR="#B22222">// seconds, though it is possible to construct test cases that
</FONT></I><I><FONT COLOR="#B22222">// achieve the worst-case.
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// Running time:
</FONT></I><I><FONT COLOR="#B22222">// O(|V|^3)
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// INPUT:
</FONT></I><I><FONT COLOR="#B22222">// - graph, constructed using AddEdge()
</FONT></I><I><FONT COLOR="#B22222">// - source
</FONT></I><I><FONT COLOR="#B22222">// - sink
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// OUTPUT:
</FONT></I><I><FONT COLOR="#B22222">// - maximum flow value
</FONT></I><I><FONT COLOR="#B22222">// - To obtain the actual flow values, look at all edges with
</FONT></I><I><FONT COLOR="#B22222">// capacity > 0 (zero capacity edges are residual edges).
</FONT></I>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><cmath></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><vector></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><iostream></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><queue></FONT></B>
using namespace std;
<B><FONT COLOR="#228B22">typedef</FONT></B> <B><FONT COLOR="#228B22">long</FONT></B> <B><FONT COLOR="#228B22">long</FONT></B> LL;
<B><FONT COLOR="#228B22">struct</FONT></B> Edge {
<B><FONT COLOR="#228B22">int</FONT></B> from, to, cap, flow, index;
Edge(<B><FONT COLOR="#228B22">int</FONT></B> from, <B><FONT COLOR="#228B22">int</FONT></B> to, <B><FONT COLOR="#228B22">int</FONT></B> cap, <B><FONT COLOR="#228B22">int</FONT></B> flow, <B><FONT COLOR="#228B22">int</FONT></B> index) :
from(from), to(to), cap(cap), flow(flow), index(index) {}
};
<B><FONT COLOR="#228B22">struct</FONT></B> PushRelabel {
<B><FONT COLOR="#228B22">int</FONT></B> N;
vector<vector<Edge> > G;
vector<LL> excess;
vector<<B><FONT COLOR="#228B22">int</FONT></B>> dist, active, count;
queue<<B><FONT COLOR="#228B22">int</FONT></B>> Q;
PushRelabel(<B><FONT COLOR="#228B22">int</FONT></B> N) : N(N), G(N), excess(N), dist(N), active(N), count(2*N) {}
<B><FONT COLOR="#228B22">void</FONT></B> AddEdge(<B><FONT COLOR="#228B22">int</FONT></B> from, <B><FONT COLOR="#228B22">int</FONT></B> to, <B><FONT COLOR="#228B22">int</FONT></B> cap) {
G[from].push_back(Edge(from, to, cap, 0, G[to].size()));
<B><FONT COLOR="#A020F0">if</FONT></B> (from == to) G[from].back().index++;
G[to].push_back(Edge(to, from, 0, 0, G[from].size() - 1));
}
<B><FONT COLOR="#228B22">void</FONT></B> Enqueue(<B><FONT COLOR="#228B22">int</FONT></B> v) {
<B><FONT COLOR="#A020F0">if</FONT></B> (!active[v] && excess[v] > 0) { active[v] = true; Q.push(v); }
}
<B><FONT COLOR="#228B22">void</FONT></B> Push(Edge &e) {
<B><FONT COLOR="#228B22">int</FONT></B> amt = <B><FONT COLOR="#228B22">int</FONT></B>(min(excess[e.from], LL(e.cap - e.flow)));
<B><FONT COLOR="#A020F0">if</FONT></B> (dist[e.from] <= dist[e.to] || amt == 0) <B><FONT COLOR="#A020F0">return</FONT></B>;
e.flow += amt;
G[e.to][e.index].flow -= amt;
excess[e.to] += amt;
excess[e.from] -= amt;
Enqueue(e.to);
}
<B><FONT COLOR="#228B22">void</FONT></B> Gap(<B><FONT COLOR="#228B22">int</FONT></B> k) {
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> v = 0; v < N; v++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (dist[v] < k) <B><FONT COLOR="#A020F0">continue</FONT></B>;
count[dist[v]]--;
dist[v] = max(dist[v], N+1);
count[dist[v]]++;
Enqueue(v);
}
}
<B><FONT COLOR="#228B22">void</FONT></B> Relabel(<B><FONT COLOR="#228B22">int</FONT></B> v) {
count[dist[v]]--;
dist[v] = 2*N;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < G[v].size(); i++)
<B><FONT COLOR="#A020F0">if</FONT></B> (G[v][i].cap - G[v][i].flow > 0)
dist[v] = min(dist[v], dist[G[v][i].to] + 1);
count[dist[v]]++;
Enqueue(v);
}
<B><FONT COLOR="#228B22">void</FONT></B> Discharge(<B><FONT COLOR="#228B22">int</FONT></B> v) {
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; excess[v] > 0 && i < G[v].size(); i++) Push(G[v][i]);
<B><FONT COLOR="#A020F0">if</FONT></B> (excess[v] > 0) {
<B><FONT COLOR="#A020F0">if</FONT></B> (count[dist[v]] == 1)
Gap(dist[v]);
<B><FONT COLOR="#A020F0">else</FONT></B>
Relabel(v);
}
}
LL GetMaxFlow(<B><FONT COLOR="#228B22">int</FONT></B> s, <B><FONT COLOR="#228B22">int</FONT></B> t) {
count[0] = N-1;
count[N] = 1;
dist[s] = N;
active[s] = active[t] = true;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < G[s].size(); i++) {
excess[s] += G[s][i].cap;
Push(G[s][i]);
}
<B><FONT COLOR="#A020F0">while</FONT></B> (!Q.empty()) {
<B><FONT COLOR="#228B22">int</FONT></B> v = Q.front();
Q.pop();
active[v] = false;
Discharge(v);
}
LL totflow = 0;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < G[s].size(); i++) totflow += G[s][i].flow;
<B><FONT COLOR="#A020F0">return</FONT></B> totflow;
}
};
<I><FONT COLOR="#B22222">// BEGIN CUT
</FONT></I><I><FONT COLOR="#B22222">// The following code solves SPOJ problem #4110: Fast Maximum Flow (FASTFLOW)
</FONT></I>
<B><FONT COLOR="#228B22">int</FONT></B> <B><FONT COLOR="#0000FF">main</FONT></B>() {
<B><FONT COLOR="#228B22">int</FONT></B> n, m;
scanf(<B><FONT COLOR="#BC8F8F">"%d%d"</FONT></B>, &n, &m);
PushRelabel pr(n);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < m; i++) {
<B><FONT COLOR="#228B22">int</FONT></B> a, b, c;
scanf(<B><FONT COLOR="#BC8F8F">"%d%d%d"</FONT></B>, &a, &b, &c);
<B><FONT COLOR="#A020F0">if</FONT></B> (a == b) <B><FONT COLOR="#A020F0">continue</FONT></B>;
pr.AddEdge(a-1, b-1, c);
pr.AddEdge(b-1, a-1, c);
}
printf(<B><FONT COLOR="#BC8F8F">"%Ld\n"</FONT></B>, pr.GetMaxFlow(0, n-1));
<B><FONT COLOR="#A020F0">return</FONT></B> 0;
}
<I><FONT COLOR="#B22222">// END CUT
</FONT></I></PRE>
<HR>
<A NAME="file4">
<H1>code/MinCostMatching.cc 4/27</H1>
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<PRE>
<I><FONT COLOR="#B22222">//////////////////////////////////////////////////////////////////////
</FONT></I><I><FONT COLOR="#B22222">// Min cost bipartite matching via shortest augmenting paths
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// This is an O(n^3) implementation of a shortest augmenting path
</FONT></I><I><FONT COLOR="#B22222">// algorithm for finding min cost perfect matchings in dense
</FONT></I><I><FONT COLOR="#B22222">// graphs. In practice, it solves 1000x1000 problems in around 1
</FONT></I><I><FONT COLOR="#B22222">// second.
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// cost[i][j] = cost for pairing left node i with right node j
</FONT></I><I><FONT COLOR="#B22222">// Lmate[i] = index of right node that left node i pairs with
</FONT></I><I><FONT COLOR="#B22222">// Rmate[j] = index of left node that right node j pairs with
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// The values in cost[i][j] may be positive or negative. To perform
</FONT></I><I><FONT COLOR="#B22222">// maximization, simply negate the cost[][] matrix.
</FONT></I><I><FONT COLOR="#B22222">//////////////////////////////////////////////////////////////////////
</FONT></I>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><algorithm></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><cstdio></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><cmath></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><vector></FONT></B>
using namespace std;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<<B><FONT COLOR="#228B22">double</FONT></B>> VD;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<VD> VVD;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<<B><FONT COLOR="#228B22">int</FONT></B>> VI;
<B><FONT COLOR="#228B22">double</FONT></B> <B><FONT COLOR="#0000FF">MinCostMatching</FONT></B>(<B><FONT COLOR="#228B22">const</FONT></B> VVD &cost, VI &Lmate, VI &Rmate) {
<B><FONT COLOR="#228B22">int</FONT></B> n = <B><FONT COLOR="#228B22">int</FONT></B>(cost.size());
<I><FONT COLOR="#B22222">// construct dual feasible solution
</FONT></I> VD u(n);
VD v(n);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < n; i++) {
u[i] = cost[i][0];
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 1; j < n; j++) u[i] = min(u[i], cost[i][j]);
}
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 0; j < n; j++) {
v[j] = cost[0][j] - u[0];
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 1; i < n; i++) v[j] = min(v[j], cost[i][j] - u[i]);
}
<I><FONT COLOR="#B22222">// construct primal solution satisfying complementary slackness
</FONT></I> Lmate = VI(n, -1);
Rmate = VI(n, -1);
<B><FONT COLOR="#228B22">int</FONT></B> mated = 0;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < n; i++) {
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 0; j < n; j++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (Rmate[j] != -1) <B><FONT COLOR="#A020F0">continue</FONT></B>;
<B><FONT COLOR="#A020F0">if</FONT></B> (fabs(cost[i][j] - u[i] - v[j]) < 1e-10) {
Lmate[i] = j;
Rmate[j] = i;
mated++;
<B><FONT COLOR="#A020F0">break</FONT></B>;
}
}
}
VD dist(n);
VI dad(n);
VI seen(n);
<I><FONT COLOR="#B22222">// repeat until primal solution is feasible
</FONT></I> <B><FONT COLOR="#A020F0">while</FONT></B> (mated < n) {
<I><FONT COLOR="#B22222">// find an unmatched left node
</FONT></I> <B><FONT COLOR="#228B22">int</FONT></B> s = 0;
<B><FONT COLOR="#A020F0">while</FONT></B> (Lmate[s] != -1) s++;
<I><FONT COLOR="#B22222">// initialize Dijkstra
</FONT></I> fill(dad.begin(), dad.end(), -1);
fill(seen.begin(), seen.end(), 0);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> k = 0; k < n; k++)
dist[k] = cost[s][k] - u[s] - v[k];
<B><FONT COLOR="#228B22">int</FONT></B> j = 0;
<B><FONT COLOR="#A020F0">while</FONT></B> (true) {
<I><FONT COLOR="#B22222">// find closest
</FONT></I> j = -1;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> k = 0; k < n; k++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (seen[k]) <B><FONT COLOR="#A020F0">continue</FONT></B>;
<B><FONT COLOR="#A020F0">if</FONT></B> (j == -1 || dist[k] < dist[j]) j = k;
}
seen[j] = 1;
<I><FONT COLOR="#B22222">// termination condition
</FONT></I> <B><FONT COLOR="#A020F0">if</FONT></B> (Rmate[j] == -1) <B><FONT COLOR="#A020F0">break</FONT></B>;
<I><FONT COLOR="#B22222">// relax neighbors
</FONT></I> <B><FONT COLOR="#228B22">const</FONT></B> <B><FONT COLOR="#228B22">int</FONT></B> i = Rmate[j];
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> k = 0; k < n; k++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (seen[k]) <B><FONT COLOR="#A020F0">continue</FONT></B>;
<B><FONT COLOR="#228B22">const</FONT></B> <B><FONT COLOR="#228B22">double</FONT></B> new_dist = dist[j] + cost[i][k] - u[i] - v[k];
<B><FONT COLOR="#A020F0">if</FONT></B> (dist[k] > new_dist) {
dist[k] = new_dist;
dad[k] = j;
}
}
}
<I><FONT COLOR="#B22222">// update dual variables
</FONT></I> <B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> k = 0; k < n; k++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (k == j || !seen[k]) <B><FONT COLOR="#A020F0">continue</FONT></B>;
<B><FONT COLOR="#228B22">const</FONT></B> <B><FONT COLOR="#228B22">int</FONT></B> i = Rmate[k];
v[k] += dist[k] - dist[j];
u[i] -= dist[k] - dist[j];
}
u[s] += dist[j];
<I><FONT COLOR="#B22222">// augment along path
</FONT></I> <B><FONT COLOR="#A020F0">while</FONT></B> (dad[j] >= 0) {
<B><FONT COLOR="#228B22">const</FONT></B> <B><FONT COLOR="#228B22">int</FONT></B> d = dad[j];
Rmate[j] = Rmate[d];
Lmate[Rmate[j]] = j;
j = d;
}
Rmate[j] = s;
Lmate[s] = j;
mated++;
}
<B><FONT COLOR="#228B22">double</FONT></B> value = 0;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < n; i++)
value += cost[i][Lmate[i]];
<B><FONT COLOR="#A020F0">return</FONT></B> value;
}
</PRE>
<HR>
<A NAME="file5">
<H1>code/MaxBipartiteMatching.cc 5/27</H1>
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<PRE>
<I><FONT COLOR="#B22222">// This code performs maximum bipartite matching.
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// Running time: O(|E| |V|) -- often much faster in practice
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// INPUT: w[i][j] = edge between row node i and column node j
</FONT></I><I><FONT COLOR="#B22222">// OUTPUT: mr[i] = assignment for row node i, -1 if unassigned
</FONT></I><I><FONT COLOR="#B22222">// mc[j] = assignment for column node j, -1 if unassigned
</FONT></I><I><FONT COLOR="#B22222">// function returns number of matches made
</FONT></I>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><vector></FONT></B>
using namespace std;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<<B><FONT COLOR="#228B22">int</FONT></B>> VI;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<VI> VVI;
<B><FONT COLOR="#228B22">bool</FONT></B> <B><FONT COLOR="#0000FF">FindMatch</FONT></B>(<B><FONT COLOR="#228B22">int</FONT></B> i, <B><FONT COLOR="#228B22">const</FONT></B> VVI &w, VI &mr, VI &mc, VI &seen) {
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 0; j < w[i].size(); j++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (w[i][j] && !seen[j]) {
seen[j] = true;
<B><FONT COLOR="#A020F0">if</FONT></B> (mc[j] < 0 || FindMatch(mc[j], w, mr, mc, seen)) {
mr[i] = j;
mc[j] = i;
<B><FONT COLOR="#A020F0">return</FONT></B> true;
}
}
}
<B><FONT COLOR="#A020F0">return</FONT></B> false;
}
<B><FONT COLOR="#228B22">int</FONT></B> <B><FONT COLOR="#0000FF">BipartiteMatching</FONT></B>(<B><FONT COLOR="#228B22">const</FONT></B> VVI &w, VI &mr, VI &mc) {
mr = VI(w.size(), -1);
mc = VI(w[0].size(), -1);
<B><FONT COLOR="#228B22">int</FONT></B> ct = 0;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < w.size(); i++) {
VI seen(w[0].size());
<B><FONT COLOR="#A020F0">if</FONT></B> (FindMatch(i, w, mr, mc, seen)) ct++;
}
<B><FONT COLOR="#A020F0">return</FONT></B> ct;
}
</PRE>
<HR>
<A NAME="file6">
<H1>code/MinCut.cc 6/27</H1>
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<PRE>
<I><FONT COLOR="#B22222">// Adjacency matrix implementation of Stoer-Wagner min cut algorithm.
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// Running time:
</FONT></I><I><FONT COLOR="#B22222">// O(|V|^3)
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// INPUT:
</FONT></I><I><FONT COLOR="#B22222">// - graph, constructed using AddEdge()
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// OUTPUT:
</FONT></I><I><FONT COLOR="#B22222">// - (min cut value, nodes in half of min cut)
</FONT></I>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><cmath></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><vector></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><iostream></FONT></B>
using namespace std;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<<B><FONT COLOR="#228B22">int</FONT></B>> VI;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<VI> VVI;
<B><FONT COLOR="#228B22">const</FONT></B> <B><FONT COLOR="#228B22">int</FONT></B> INF = 1000000000;
pair<<B><FONT COLOR="#228B22">int</FONT></B>, VI> GetMinCut(VVI &weights) {
<B><FONT COLOR="#228B22">int</FONT></B> N = weights.size();
VI used(N), cut, best_cut;
<B><FONT COLOR="#228B22">int</FONT></B> best_weight = -1;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> phase = N-1; phase >= 0; phase--) {
VI w = weights[0];
VI added = used;
<B><FONT COLOR="#228B22">int</FONT></B> prev, last = 0;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < phase; i++) {
prev = last;
last = -1;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 1; j < N; j++)
<B><FONT COLOR="#A020F0">if</FONT></B> (!added[j] && (last == -1 || w[j] > w[last])) last = j;
<B><FONT COLOR="#A020F0">if</FONT></B> (i == phase-1) {
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 0; j < N; j++) weights[prev][j] += weights[last][j];
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 0; j < N; j++) weights[j][prev] = weights[prev][j];
used[last] = true;
cut.push_back(last);
<B><FONT COLOR="#A020F0">if</FONT></B> (best_weight == -1 || w[last] < best_weight) {
best_cut = cut;
best_weight = w[last];
}
} <B><FONT COLOR="#A020F0">else</FONT></B> {
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 0; j < N; j++)
w[j] += weights[last][j];
added[last] = true;
}
}
}
<B><FONT COLOR="#A020F0">return</FONT></B> make_pair(best_weight, best_cut);
}
<I><FONT COLOR="#B22222">// BEGIN CUT
</FONT></I><I><FONT COLOR="#B22222">// The following code solves UVA problem #10989: Bomb, Divide and Conquer
</FONT></I><B><FONT COLOR="#228B22">int</FONT></B> <B><FONT COLOR="#0000FF">main</FONT></B>() {
<B><FONT COLOR="#228B22">int</FONT></B> N;
cin >> N;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < N; i++) {
<B><FONT COLOR="#228B22">int</FONT></B> n, m;
cin >> n >> m;
VVI weights(n, VI(n));
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 0; j < m; j++) {
<B><FONT COLOR="#228B22">int</FONT></B> a, b, c;
cin >> a >> b >> c;
weights[a-1][b-1] = weights[b-1][a-1] = c;
}
pair<<B><FONT COLOR="#228B22">int</FONT></B>, VI> res = GetMinCut(weights);
cout << <B><FONT COLOR="#BC8F8F">"Case #"</FONT></B> << i+1 << <B><FONT COLOR="#BC8F8F">": "</FONT></B> << res.first << endl;
}
}
<I><FONT COLOR="#B22222">// END CUT
</FONT></I></PRE>
<HR>
<A NAME="file7">
<H1>code/GraphCutInference.cc 7/27</H1>
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<PRE>
<I><FONT COLOR="#B22222">// Special-purpose {0,1} combinatorial optimization solver for
</FONT></I><I><FONT COLOR="#B22222">// problems of the following by a reduction to graph cuts:
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// minimize sum_i psi_i(x[i])
</FONT></I><I><FONT COLOR="#B22222">// x[1]...x[n] in {0,1} + sum_{i < j} phi_{ij}(x[i], x[j])
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// where
</FONT></I><I><FONT COLOR="#B22222">// psi_i : {0, 1} --> R
</FONT></I><I><FONT COLOR="#B22222">// phi_{ij} : {0, 1} x {0, 1} --> R
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// such that
</FONT></I><I><FONT COLOR="#B22222">// phi_{ij}(0,0) + phi_{ij}(1,1) <= phi_{ij}(0,1) + phi_{ij}(1,0) (*)
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// This can also be used to solve maximization problems where the
</FONT></I><I><FONT COLOR="#B22222">// direction of the inequality in (*) is reversed.
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// INPUT: phi -- a matrix such that phi[i][j][u][v] = phi_{ij}(u, v)
</FONT></I><I><FONT COLOR="#B22222">// psi -- a matrix such that psi[i][u] = psi_i(u)
</FONT></I><I><FONT COLOR="#B22222">// x -- a vector where the optimal solution will be stored
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// OUTPUT: value of the optimal solution
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// To use this code, create a GraphCutInference object, and call the
</FONT></I><I><FONT COLOR="#B22222">// DoInference() method. To perform maximization instead of minimization,
</FONT></I><I><FONT COLOR="#B22222">// ensure that #define MAXIMIZATION is enabled.
</FONT></I>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><vector></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><iostream></FONT></B>
using namespace std;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<<B><FONT COLOR="#228B22">int</FONT></B>> VI;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<VI> VVI;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<VVI> VVVI;
<B><FONT COLOR="#228B22">typedef</FONT></B> vector<VVVI> VVVVI;
<B><FONT COLOR="#228B22">const</FONT></B> <B><FONT COLOR="#228B22">int</FONT></B> INF = 1000000000;
<I><FONT COLOR="#B22222">// comment out following line for minimization
</FONT></I>#<B><FONT COLOR="#5F9EA0">define</FONT></B> <FONT COLOR="#B8860B">MAXIMIZATION</FONT>
<B><FONT COLOR="#228B22">struct</FONT></B> GraphCutInference {
<B><FONT COLOR="#228B22">int</FONT></B> N;
VVI cap, flow;
VI reached;
<B><FONT COLOR="#228B22">int</FONT></B> Augment(<B><FONT COLOR="#228B22">int</FONT></B> s, <B><FONT COLOR="#228B22">int</FONT></B> t, <B><FONT COLOR="#228B22">int</FONT></B> a) {
reached[s] = 1;
<B><FONT COLOR="#A020F0">if</FONT></B> (s == t) <B><FONT COLOR="#A020F0">return</FONT></B> a;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> k = 0; k < N; k++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (reached[k]) <B><FONT COLOR="#A020F0">continue</FONT></B>;
<B><FONT COLOR="#A020F0">if</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> aa = min(a, cap[s][k] - flow[s][k])) {
<B><FONT COLOR="#A020F0">if</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> b = Augment(k, t, aa)) {
flow[s][k] += b;
flow[k][s] -= b;
<B><FONT COLOR="#A020F0">return</FONT></B> b;
}
}
}
<B><FONT COLOR="#A020F0">return</FONT></B> 0;
}
<B><FONT COLOR="#228B22">int</FONT></B> GetMaxFlow(<B><FONT COLOR="#228B22">int</FONT></B> s, <B><FONT COLOR="#228B22">int</FONT></B> t) {
N = cap.size();
flow = VVI(N, VI(N));
reached = VI(N);
<B><FONT COLOR="#228B22">int</FONT></B> totflow = 0;
<B><FONT COLOR="#A020F0">while</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> amt = Augment(s, t, INF)) {
totflow += amt;
fill(reached.begin(), reached.end(), 0);
}
<B><FONT COLOR="#A020F0">return</FONT></B> totflow;
}
<B><FONT COLOR="#228B22">int</FONT></B> DoInference(<B><FONT COLOR="#228B22">const</FONT></B> VVVVI &phi, <B><FONT COLOR="#228B22">const</FONT></B> VVI &psi, VI &x) {
<B><FONT COLOR="#228B22">int</FONT></B> M = phi.size();
cap = VVI(M+2, VI(M+2));
VI b(M);
<B><FONT COLOR="#228B22">int</FONT></B> c = 0;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < M; i++) {
b[i] += psi[i][1] - psi[i][0];
c += psi[i][0];
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = 0; j < i; j++)
b[i] += phi[i][j][1][1] - phi[i][j][0][1];
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = i+1; j < M; j++) {
cap[i][j] = phi[i][j][0][1] + phi[i][j][1][0] - phi[i][j][0][0] - phi[i][j][1][1];
b[i] += phi[i][j][1][0] - phi[i][j][0][0];
c += phi[i][j][0][0];
}
}
#<B><FONT COLOR="#5F9EA0">ifdef</FONT></B> <FONT COLOR="#B8860B">MAXIMIZATION</FONT>
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < M; i++) {
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> j = i+1; j < M; j++)
cap[i][j] *= -1;
b[i] *= -1;
}
c *= -1;
#<B><FONT COLOR="#5F9EA0">endif
</FONT></B>
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < M; i++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (b[i] >= 0) {
cap[M][i] = b[i];
} <B><FONT COLOR="#A020F0">else</FONT></B> {
cap[i][M+1] = -b[i];
c += b[i];
}
}
<B><FONT COLOR="#228B22">int</FONT></B> score = GetMaxFlow(M, M+1);
fill(reached.begin(), reached.end(), 0);
Augment(M, M+1, INF);
x = VI(M);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < M; i++) x[i] = reached[i] ? 0 : 1;
score += c;
#<B><FONT COLOR="#5F9EA0">ifdef</FONT></B> <FONT COLOR="#B8860B">MAXIMIZATION</FONT>
score *= -1;
#<B><FONT COLOR="#5F9EA0">endif
</FONT></B>
<B><FONT COLOR="#A020F0">return</FONT></B> score;
}
};
<B><FONT COLOR="#228B22">int</FONT></B> <B><FONT COLOR="#0000FF">main</FONT></B>() {
<I><FONT COLOR="#B22222">// solver for "Cat vs. Dog" from NWERC 2008
</FONT></I>
<B><FONT COLOR="#228B22">int</FONT></B> numcases;
cin >> numcases;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> caseno = 0; caseno < numcases; caseno++) {
<B><FONT COLOR="#228B22">int</FONT></B> c, d, v;
cin >> c >> d >> v;
VVVVI phi(c+d, VVVI(c+d, VVI(2, VI(2))));
VVI psi(c+d, VI(2));
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < v; i++) {
<B><FONT COLOR="#228B22">char</FONT></B> p, q;
<B><FONT COLOR="#228B22">int</FONT></B> u, v;
cin >> p >> u >> q >> v;
u--; v--;
<B><FONT COLOR="#A020F0">if</FONT></B> (p == <B><FONT COLOR="#BC8F8F">'C'</FONT></B>) {
phi[u][c+v][0][0]++;
phi[c+v][u][0][0]++;
} <B><FONT COLOR="#A020F0">else</FONT></B> {
phi[v][c+u][1][1]++;
phi[c+u][v][1][1]++;
}
}
GraphCutInference graph;
VI x;
cout << graph.DoInference(phi, psi, x) << endl;
}
<B><FONT COLOR="#A020F0">return</FONT></B> 0;
}
</PRE>
<HR>
<A NAME="file8">
<H1>code/ConvexHull.cc 8/27</H1>
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<PRE>
<I><FONT COLOR="#B22222">// Compute the 2D convex hull of a set of points using the monotone chain
</FONT></I><I><FONT COLOR="#B22222">// algorithm. Eliminate redundant points from the hull if REMOVE_REDUNDANT is
</FONT></I><I><FONT COLOR="#B22222">// #defined.
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// Running time: O(n log n)
</FONT></I><I><FONT COLOR="#B22222">//
</FONT></I><I><FONT COLOR="#B22222">// INPUT: a vector of input points, unordered.
</FONT></I><I><FONT COLOR="#B22222">// OUTPUT: a vector of points in the convex hull, counterclockwise, starting
</FONT></I><I><FONT COLOR="#B22222">// with bottommost/leftmost point
</FONT></I>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><cstdio></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><cassert></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><vector></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><algorithm></FONT></B>
#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><cmath></FONT></B>
<I><FONT COLOR="#B22222">// BEGIN CUT
</FONT></I>#<B><FONT COLOR="#5F9EA0">include</FONT></B> <B><FONT COLOR="#BC8F8F"><map></FONT></B>
<I><FONT COLOR="#B22222">// END CUT
</FONT></I>
using namespace std;
#<B><FONT COLOR="#5F9EA0">define</FONT></B> <FONT COLOR="#B8860B">REMOVE_REDUNDANT</FONT>
<B><FONT COLOR="#228B22">typedef</FONT></B> <B><FONT COLOR="#228B22">double</FONT></B> T;
<B><FONT COLOR="#228B22">const</FONT></B> T EPS = 1e-7;
<B><FONT COLOR="#228B22">struct</FONT></B> PT {
T x, y;
PT() {}
PT(T x, T y) : x(x), y(y) {}
<B><FONT COLOR="#228B22">bool</FONT></B> <B><FONT COLOR="#A020F0">operator</FONT></B><(<B><FONT COLOR="#228B22">const</FONT></B> PT &rhs) <B><FONT COLOR="#228B22">const</FONT></B> { <B><FONT COLOR="#A020F0">return</FONT></B> make_pair(y,x) < make_pair(rhs.y,rhs.x); }
<B><FONT COLOR="#228B22">bool</FONT></B> <B><FONT COLOR="#A020F0">operator</FONT></B>==(<B><FONT COLOR="#228B22">const</FONT></B> PT &rhs) <B><FONT COLOR="#228B22">const</FONT></B> { <B><FONT COLOR="#A020F0">return</FONT></B> make_pair(y,x) == make_pair(rhs.y,rhs.x); }
};
T cross(PT p, PT q) { <B><FONT COLOR="#A020F0">return</FONT></B> p.x*q.y-p.y*q.x; }
T area2(PT a, PT b, PT c) { <B><FONT COLOR="#A020F0">return</FONT></B> cross(a,b) + cross(b,c) + cross(c,a); }
#<B><FONT COLOR="#5F9EA0">ifdef</FONT></B> <FONT COLOR="#B8860B">REMOVE_REDUNDANT</FONT>
<B><FONT COLOR="#228B22">bool</FONT></B> <B><FONT COLOR="#0000FF">between</FONT></B>(<B><FONT COLOR="#228B22">const</FONT></B> PT &a, <B><FONT COLOR="#228B22">const</FONT></B> PT &b, <B><FONT COLOR="#228B22">const</FONT></B> PT &c) {
<B><FONT COLOR="#A020F0">return</FONT></B> (fabs(area2(a,b,c)) < EPS && (a.x-b.x)*(c.x-b.x) <= 0 && (a.y-b.y)*(c.y-b.y) <= 0);
}
#<B><FONT COLOR="#5F9EA0">endif
</FONT></B>
<B><FONT COLOR="#228B22">void</FONT></B> <B><FONT COLOR="#0000FF">ConvexHull</FONT></B>(vector<PT> &pts) {
sort(pts.begin(), pts.end());
pts.erase(unique(pts.begin(), pts.end()), pts.end());
vector<PT> up, dn;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < pts.size(); i++) {
<B><FONT COLOR="#A020F0">while</FONT></B> (up.size() > 1 && area2(up[up.size()-2], up.back(), pts[i]) >= 0) up.pop_back();
<B><FONT COLOR="#A020F0">while</FONT></B> (dn.size() > 1 && area2(dn[dn.size()-2], dn.back(), pts[i]) <= 0) dn.pop_back();
up.push_back(pts[i]);
dn.push_back(pts[i]);
}
pts = dn;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = (<B><FONT COLOR="#228B22">int</FONT></B>) up.size() - 2; i >= 1; i--) pts.push_back(up[i]);
#<B><FONT COLOR="#5F9EA0">ifdef</FONT></B> <FONT COLOR="#B8860B">REMOVE_REDUNDANT</FONT>
<B><FONT COLOR="#A020F0">if</FONT></B> (pts.size() <= 2) <B><FONT COLOR="#A020F0">return</FONT></B>;
dn.clear();
dn.push_back(pts[0]);
dn.push_back(pts[1]);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 2; i < pts.size(); i++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (between(dn[dn.size()-2], dn[dn.size()-1], pts[i])) dn.pop_back();
dn.push_back(pts[i]);
}
<B><FONT COLOR="#A020F0">if</FONT></B> (dn.size() >= 3 && between(dn.back(), dn[0], dn[1])) {
dn[0] = dn.back();
dn.pop_back();
}
pts = dn;
#<B><FONT COLOR="#5F9EA0">endif
</FONT></B>
}
<I><FONT COLOR="#B22222">// BEGIN CUT
</FONT></I><I><FONT COLOR="#B22222">// The following code solves SPOJ problem #26: Build the Fence (BSHEEP)
</FONT></I>
<B><FONT COLOR="#228B22">int</FONT></B> <B><FONT COLOR="#0000FF">main</FONT></B>() {
<B><FONT COLOR="#228B22">int</FONT></B> t;
scanf(<B><FONT COLOR="#BC8F8F">"%d"</FONT></B>, &t);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> caseno = 0; caseno < t; caseno++) {
<B><FONT COLOR="#228B22">int</FONT></B> n;
scanf(<B><FONT COLOR="#BC8F8F">"%d"</FONT></B>, &n);
vector<PT> v(n);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < n; i++) scanf(<B><FONT COLOR="#BC8F8F">"%lf%lf"</FONT></B>, &v[i].x, &v[i].y);
vector<PT> h(v);
map<PT,<B><FONT COLOR="#228B22">int</FONT></B>> index;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = n-1; i >= 0; i--) index[v[i]] = i+1;
ConvexHull(h);
<B><FONT COLOR="#228B22">double</FONT></B> len = 0;
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < h.size(); i++) {
<B><FONT COLOR="#228B22">double</FONT></B> dx = h[i].x - h[(i+1)%h.size()].x;
<B><FONT COLOR="#228B22">double</FONT></B> dy = h[i].y - h[(i+1)%h.size()].y;
len += sqrt(dx*dx+dy*dy);
}
<B><FONT COLOR="#A020F0">if</FONT></B> (caseno > 0) printf(<B><FONT COLOR="#BC8F8F">"\n"</FONT></B>);
printf(<B><FONT COLOR="#BC8F8F">"%.2f\n"</FONT></B>, len);
<B><FONT COLOR="#A020F0">for</FONT></B> (<B><FONT COLOR="#228B22">int</FONT></B> i = 0; i < h.size(); i++) {
<B><FONT COLOR="#A020F0">if</FONT></B> (i > 0) printf(<B><FONT COLOR="#BC8F8F">" "</FONT></B>);
printf(<B><FONT COLOR="#BC8F8F">"%d"</FONT></B>, index[h[i]]);
}
printf(<B><FONT COLOR="#BC8F8F">"\n"</FONT></B>);
}