有p门的课,每门课都有若干学生,现在要为每个课程分配一名课代表,每个学生只能担任一门课的课代表,如果每个课都能找到课代表,则输出”YES”,否则”NO”。
二分图最大匹配.
看看是不是完美匹配即可.
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#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x+1)
#define oi(x) printf("%d",x)
#define ol(x) printf("%lld",x)
#define oc putchar(' ')
#define os(x) printf(x)
#define all(x) x.begin(),x.end()
#define Open() freopen("F:\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 400;
int p,n,pre[N+10],Try[N+10];
vector <int> g[N+10];
bool hungary(int x){
int len = g[x].size();
rep1(i,0,len-1){
int y = g[x][i];
if (!Try[y]){
Try[y] = 1;
if (pre[y]==-1 || hungary(pre[y])){
pre[y] = x;
return true;
}
}
}
return false;
}
int main(){
//Open();
//Close();
int T;
ri(T);
while (T--){
rep1(i,1,N) g[i].clear();
ri(p),ri(n);
rep1(i,1,p){
int cnt;
ri(cnt);
rep1(j,1,cnt){
int x;
ri(x);
g[x].pb(i + n);
}
}
int ans = 0;
ms(pre,255);
rep1(i,1,n){
ms(Try,0);
if (hungary(i)) ans++;
}
if (ans == p)
puts("YES");
else
puts("NO");
}
return 0;
}