题意：有P门课程，N个学生，每门课程有一些学生选读，每个学生选读一些课程，问能否选出P个学生组成一个委员会，使得每个学生代言一门课程（他必需选读其代言的课程），每门课程都被一个学生代言(1 <= P <= 100，1 <= N <= 300) 。
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1083
——>>第一次自己想出的网络流。。。虽然是水题，但也开心死死。。。
建图：设超级源S，S到每门课程连一条边，容量为1；每门课程向其选读的学生各连一条边，容量为1；每个学生向超级汇连一条边，容量为1。
这样，只要求一次最大流，判断其是否为满流P就好。。。
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cstring>using namespace std;const int maxn = 400 + 10;
const int maxm = 60800 + 10;
const int INF = 0x3f3f3f3f;int head[maxn], nxt[maxm], ecnt, v[maxm], flow[maxm], cap[maxm];
bool flag[maxn];struct Dinic{int m, s, t;int d[maxn], cur[maxn];bool vis[maxn];Dinic(){memset(head, -1, sizeof(head));ecnt = 0;}void addEdge(int uu, int vv, int ca){v[ecnt] = vv; cap[ecnt] = ca; flow[ecnt] = 0; nxt[ecnt] = head[uu]; head[uu] = ecnt; ecnt++;v[ecnt] = uu; cap[ecnt] = 0; flow[ecnt] = 0; nxt[ecnt] = head[vv]; head[vv] = ecnt; ecnt++;}bool bfs(){d[s] = 0;memset(vis, 0, sizeof(vis));queue<int> qu;qu.push(s);vis[s] = 1;while(!qu.empty()){int u = qu.front(); qu.pop();for(int e = head[u]; e != -1; e = nxt[e]){if(!vis[v[e]] && cap[e] > flow[e]){d[v[e]] = d[u] + 1;vis[v[e]] = 1;qu.push(v[e]);}}}return vis[t];}int dfs(int u, int a){if(u == t || a == 0) return a;int f, Flow = 0;for(int e = cur[u]; e != -1; e = nxt[e]){cur[u] = e;if(d[v[e]] == d[u] + 1 && (f = dfs(v[e], min(a, cap[e]-flow[e]))) > 0){flow[e] += f;flow[e^1] -= f;Flow += f;a -= f;if(!a) break;}}return Flow;}int Maxflow(int s, int t){this->s = s;this->t = t;int Flow = 0;while(bfs()){memcpy(cur, head, sizeof(head));Flow += dfs(s, INF);}return Flow;}};int main()
{int T, P, N, S, cnt;scanf("%d", &T);while(T--){Dinic din;scanf("%d%d", &P, &N);for(int i = 1; i <= P; i++){din.addEdge(0, i, 1);scanf("%d", &cnt);for(int j = 1; j <= cnt; j++){scanf("%d", &S);din.addEdge(i, P+S, 1);}}for(int i = 1; i <= N; i++) din.addEdge(P+i, P+N+1, 1);if(din.Maxflow(0, P+N+1) == P) puts("YES");else puts("NO");}return 0;
}