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本文介绍了一个集合操作查询问题的算法实现。该问题涉及处理多个可能包含重复元素的集合，并通过高效的位运算方法来判断任意两个元素是否至少在一个共同的集合中出现。文章详细解释了输入输出格式，并给出了一段具体的C++代码示例。

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;
}

转载于:https://www.cnblogs.com/water-full/p/4815063.html