题目传送门

思路

这是个板子题,二分图的最小点覆盖

#include <iostream>
using namespace std;
int e[500][500], n, m, match[500], ans, book[500], cow[500];
int dfs(int u){
    for (int i = 1; i <= m; i++){
        if (e[u][i] == 1 && book[i] == 0){
            book[i] = 1;
            if (match[i] == 0 || dfs(match[i]) == 1){
                match[i] = u;
                return 1;
            }
        }
    }
    return 0;
}
int main(){
    int t, x, p;
    cin >> n >> m >> p;
    for (int i = 1; i <= p; i++){
        cin >> t >> x;
        e[t][x] = 1;
    }
    for (int i = 1; i <= n; i++){
        memset(book, 0, sizeof(book));
        if (dfs(i) == 1) ans ++;
    }
    cout << ans << endl;
    return 0;
}

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