POJ 1469 COURSES 二分圖最大匹配

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就是判斷一下是不是每一個課程都能找到自己的代表人，做一遍最大匹配看看匹配數是否等於p即可

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <climits>#include <string>#include <iostream>#include <map>#include <cstdlib>#include <list>#include <set>#include <queue>#include <stack>using namespace std;typedef long long LL;const int maxn = 400;int p,n,bx[maxn],by[maxn],vis[maxn];bool g[maxn][maxn];int dfs(int now) {    for(int i = 1;i <= n;i++) if(g[now][i] && !vis[i]) {        vis[i] = true;        if(!by[i] || dfs(by[i])) {            bx[now] = i; by[i] = now;            return 1;        }    }    return 0;}void solve() {    memset(bx,0,sizeof(bx));    memset(by,0,sizeof(by));    //為每一門課程尋找代表    int ans = 0;    for(int i = 1;i <= p;i++) {        if(!bx[i]) {            memset(vis,0,sizeof(vis));            ans += dfs(i);        }    }    if(ans == p) puts("YES");    else puts("NO");}int main() {    int T; scanf("%d",&T);    while(T--) {        scanf("%d%d",&p,&n);        memset(g,0,sizeof(g));        for(int i = 1;i <= p;i++) {            int cnt; scanf("%d",&cnt);            for(int j = 1;j <= cnt;j++) {                int tmp; scanf("%d",&tmp);                g[i][tmp] = true;            }        }        solve();    }    return 0;}