题目大意：
有n1头公牛和n2头母牛，给出公母之间的m对配对关系，求最大匹配数。
数据范围： 1 <= n1, n2 <= 50000, m <= 150000
注意：该模板存边使用的是前向星,且二部图左右两边的点集编号均从1开始
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
const int maxn = 5e4+5;
const int maxe = 15e4+5;//最大关系数
const int INF = 1<<28;
int tol, head[maxn];
int Mx[maxn], My[maxn], Nx, Ny;
//Nx左边点集合的个数,标号从1开始
//Ny左边点集合的个数,标号从1开始
int dx[maxn], dy[maxn], dis;
bool vst[maxn*2];
struct Edge
{int from, to, nxt;
}edge[maxe];
void init()
{tol = 0;memset(head, -1, sizeof(head));
}
void addedge(int u, int v)
{edge[tol].from = u;edge[tol].to = v;edge[tol].nxt = head[u];head[u] = tol++;
}
bool searchP()
{queue<int> Q;dis = INF;memset(dx, -1, sizeof(dx));memset(dy, -1, sizeof(dy));for(int i = 1; i <= Nx; i++)if(Mx[i] == -1){Q.push(i);dx[i] = 0;}while(!Q.empty()){int u = Q.front();Q.pop();if(dx[u] > dis) break;for(int i = head[u]; i != -1; i = edge[i].nxt){int v = edge[i].to;if(dy[v] != -1) continue;dy[v] = dx[u] + 1;if(My[v] == -1)dis = dy[v];else{dx[My[v]] = dy[v]+1;Q.push(My[v]);}}}return dis!=INF;
}
bool DFS(int u)
{for(int i = head[u]; i != -1; i = edge[i].nxt){int v = edge[i].to;if(!vst[v] && dy[v] == dx[u]+1){vst[v]=1;if(My[v]!=-1&&dy[v]==dis) continue;if(My[v]==-1||DFS(My[v])){My[v]=u;Mx[u]=v;return true;}}}return false;
}
int MaxMatch()
{int res = 0;memset(Mx, -1, sizeof(Mx));memset(My, -1, sizeof(My));while(searchP()){memset(vst, false, sizeof(vst));for(int i = 1; i <= Nx; i++)if(Mx[i] == -1 && DFS(i))res++;}return res;
}
int main()
{int n, m, tol_edge, u, v;//n左边点的数量,m右边点的数量,tol_edge关系数while(scanf("%d%d%d", &n, &m, &tol_edge) != EOF){init();for(int i = 0; i < tol_edge; i++){scanf("%d%d", &u, &v);addedge(u, v);}Nx = n;Ny = m;printf("%d\n", MaxMatch());}return 0;
}