題意:有P門課程,N個學生,每門課程有一些學生選讀,每個學生選讀一些課程,問能否選出P個學生組成一個委員會,使得每個學生代言一門課程(他必需選讀其代言的課程),每門課程都被一個學生代言(1 <= P <= 100,1 <= N <= 300) 。 ——>>第一次自己想出的網絡流。。。雖然是水題,但也開心死死。。。 建圖:設超級源S,S到每門課程連一條邊,容量為1;每門課程向其選讀的學生各連一條邊,容量為1;每個學生向超級匯連一條邊,容量為1。 這樣,只要求一次最大流,判斷其是否為滿流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;
}