P3386 【模板】二分图匹配
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 2005, inf = 0x3f3f3f; 4 struct Edge { 5 int from, to, cap, flow; 6 }; 7 8 struct Dinic { 9 int n, m, s, t; 10 vector<Edge> edges; 11 vector<int> G[maxn]; 12 bool vis[maxn]; 13 int d[maxn]; 14 int cur[maxn]; 15 16 void AddEdge(int from, int to, int cap) { 17 edges.push_back((Edge){from, to, cap, 0}); 18 edges.push_back((Edge){to, from, 0, 0}); 19 m = edges.size(); 20 G[from].push_back(m-2); 21 G[to].push_back(m-1); 22 } 23 bool bfs() { 24 memset(vis, 0, sizeof(vis)); 25 queue<int> que; 26 que.push(s); 27 d[s] = 0; 28 vis[s] = true; 29 while (!que.empty()) { 30 int x = que.front(); que.pop(); 31 for (int i = 0; i < G[x].size(); ++i) { 32 Edge& e = edges[G[x][i]]; 33 if (!vis[e.to] && e.cap > e.flow) { 34 vis[e.to] = true; 35 d[e.to] = d[x] + 1; 36 que.push(e.to); 37 } 38 } 39 } 40 return vis[t]; 41 } 42 int dfs(int x, int a) { 43 if (x == t || a == 0) return a; 44 int flow = 0, f; 45 for (int& i = cur[x]; i < G[x].size(); ++i) { 46 Edge& e = edges[G[x][i]]; 47 if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0) { 48 e.flow += f; 49 edges[G[x][i]^1].flow -= f; 50 flow += f; 51 a -= f; 52 if (a == 0) break; 53 } 54 } 55 return flow; 56 } 57 int maxflow(int s, int t) { 58 this->s = s; this->t = t; 59 int flow = 0; 60 while (bfs()) { 61 memset(cur,0,sizeof(cur)); 62 flow += dfs(s,inf); 63 } 64 return flow; 65 } 66 }dinic; 67 int main() { 68 int n, m, e; scanf("%d%d%d",&n,&m,&e); 69 int s = n+m+1, t = n+m+2; 70 for (int i = 1; i <= n; ++i) { 71 dinic.AddEdge(s,i,1); 72 } 73 for (int i = 1; i <= m; ++i) { 74 dinic.AddEdge(i+n,t,1); 75 } 76 for (int i = 1; i <= e; ++i) { 77 int u, v; scanf("%d%d",&u,&v); 78 dinic.AddEdge(u,v+n,1); 79 } 80 int ans = dinic.maxflow(s,t); 81 printf("%d\n",ans); 82 return 0; 83 }
