——>>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#include #include #include 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 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; }