p3386 二分图匹配模板

https://www.luogu.org/problemnew/show/P3386

可以只做一边的匹配

#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e5 + 7;
vector<int> G[maxN];
int match[maxN];
int vis[maxN];
int n, m, e, sum;
int dfs(int u) {

    for(int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        //有路而且没被访问
        if(!vis[v]) {
            vis[v] = 1;//标记点i已经访问过

            //如果点i未被配对或者找到了新的配对
            if(match[v] == 0 || dfs(match[v])) {
                //更新配对关系
                match[v] = u;
                return 1;
            }
        }
    }
    return 0;
}

int main() {
//    freopen("testdata.in","r", stdin);
    scanf("%d %d %d", &n, &m, &e);
    for(int i = 0; i < e; i++) {
        int u, v;
        scanf("%d %d", &u, &v);
        if(u > n || v > m) continue; //u总是在左边， 只做u的匹配
//        v += 10000;
        G[u].push_back(v);//只做一边建单向边
//        G[v].push_back(u);
    }

    memset(match, 0 , sizeof(match));

    for(int i = 1; i <= n; i++){
        memset(vis, 0, sizeof(vis));
        if(dfs(i)) {
            sum++;
        }
    }
    printf("%d
", sum);


}