hdu 5506 GT and set dfs+bitset优化

GT and set

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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.

 

 

Source

BestCoder Round #60

 题意:如果n个集合的交集不为空就可以合并，问n个能合并为L个；

思路：暴力搜索，两个集合合并的话用bitset优化即可；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+10,M=1e6+10,inf=1e9+10;
const ll INF=1e18+10,mod=2147493647;
bitset<310> flag[10],a[50],temp,bit;
int n,m,ans;
void dfs(int pos)
{
    if(pos>n||ans)
    {
        ans=1;
        return;
    }
    for(int i=1;i<=m;i++)
    {
        temp=flag[i]&a[pos];
        bit=flag[i];
        if(temp.count())
        {
            flag[i]=temp;
            dfs(pos+1);
            flag[i]=bit;
        }
    }
}
void init()
{
    for(int i=1;i<=n;i++)
        a[i].reset();
    for(int i=1;i<=m;i++)
        flag[i].set();
    ans=0;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=n;i++)
        {
            int z;
            scanf("%d",&z);
            for(int j=1;j<=z;j++)
            {
                int x;
                scanf("%d",&x);
                a[i].set(x);
            }
        }
        dfs(1);
        if(ans)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}