GT and set

GT and set

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 605    Accepted Submission(s): 211

Problem Description

You are given N sets.The i−th set has Ai numbers.

You should divide the sets into L parts.

And each part should have at least one number in common.

If there is at least one solution,print YES,otherwise print NO.

 

 

Input

In the first line there is the testcase T (T≤20)

For each teatcase:

In the first line there are two numbers N and L.

In the next N lines,each line describe a set.

The first number is Ai,and then there are Ai distict numbers stand for the elements int the set.

The numbers in the set are all positive numbers and they're all not bigger than 300.

1≤N≤30，1≤L≤5，1≤Ai≤10，1≤L≤N

You'd better print the enter in the last line when you hack others.

You'd better not print space in the last of each line when you hack others.

 

 

Output

For each test print YES or NO

 

 

Sample Input

2 2 1 1 1 1 2 3 2 3 1 2 3 3 4 5 6 3 2 5 6

 

 

Sample Output

NO YES

Hint

For the second test,there are three sets:{1,2,3},{4,5,6},{2,5,6} You are asked to divide into two parts. One possible solution is to put the second and the third sets into the same part,and put the first in the other part. The second part and the third part have same number 6. Another solution is to put the first and the third sets into the same part,and put the second in the other part.

 

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 2e5 + 50;

int T,n,l,vis[1008];
vector<int>e[maxn];

inline bool check(int cur){
    for(register int i=0;i<e[cur].size();++i){
        if(vis[e[cur][i]])return true;
    }
    return false;
}

inline bool solve(int cur,int part){
    if(cur>n)return true;
    if(check(cur))return solve(cur+1,part);
    if(part>=l)return false;
    for(register int i=0;i<e[cur].size();++i){
        vis[e[cur][i]]=1;
        if(solve(cur+1,part+1))return true;
        vis[e[cur][i]]=0;
    }
    return false;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&l);
        memset(vis,0,sizeof(vis));
        for(register int i=1;i<=n;++i)e[i].clear();
        for(register int i=1,size;i<=n;++i){
            scanf("%d",&size);
            for(register int j=1,x;j<=size;++j){
                scanf("%d",&x);
                e[i].emplace_back(x);
            }
        }
        printf("%s\n",solve(1,0)?"YES":"NO");
    }
    return 0;
}