poj 2443

时间：2015-09-17 06:19:06 阅读：203 评论：0 收藏：0 [点我收藏+]

Set Operation

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 2961		Accepted: 1192

Description

You are given N sets, the i-th set (represent by S(i)) have C(i) element (Here "set" isn‘t entirely the same as the "set" defined in mathematics, and a set may contain two same element). Every element in a set is represented by a positive number from 1 to 10000. Now there are some queries need to answer. A query is to determine whether two given elements i and j belong to at least one set at the same time. In another word, you should determine if there exist a number k (1 <= k <= N) such that element i belongs to S(k) and element j also belong to S(k).

Input

First line of input contains an integer N (1 <= N <= 1000), which represents the amount of sets. Then follow N lines. Each starts with a number C(i) (1 <= C(i) <= 10000), and then C(i) numbers, which are separated with a space, follow to give the element in the set (these C(i) numbers needn‘t be different from each other). The N + 2 line contains a number Q (1 <= Q <= 200000), representing the number of queries. Then follow Q lines. Each contains a pair of number i and j (1 <= i, j <= 10000, and i may equal to j), which describe the elements need to be answer.

Output

For each query, in a single line, if there exist such a number k, print "Yes"; otherwise print "No".

Sample Input

Sample Output

Yes
Yes
No
No

Hint

The input may be large, and the I/O functions (cin/cout) of C++ language may be a little too slow for this problem.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;
int n,m;
int a[32][10005];
int main()
{
    int num,tt,q;
    while(scanf("%d",&n)!=EOF)
    {
        bool flag;
        memset(a,0,sizeof(a));
        for(int i=0;i<n;i++)
        {
            scanf("%d",&num);
            int xx,yy;
            xx=i/32;
            yy=i%32;
            for(int j=0;j<num;j++)
            {
                scanf("%d",&tt);
                a[xx][tt]|=(1<<yy);
            }
        }
        m=n/32;
        scanf("%d",&q);
        for(int i=0;i<q;i++)
        {
            int u,v;
            flag=false;
            scanf("%d%d",&u,&v);
            for(int j=0;j<=m;j++)
            {
                if(a[j][u]&a[j][v])
                {
                    flag=1;
                    break;
                }
            }
            if(flag)
                printf("Yes\n");
            else
                printf("No\n");
        }
    }
    return 0;
}

poj 2443

原文：http://www.cnblogs.com/water-full/p/4815063.html

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