POJ2443：Set Operation（bitset）

Set Operation

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 3223 Accepted: 1310

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

33 1 2 33 1 2 51 1041 31 53 51 10

Sample Output

YesYesNoNo

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.

Source

POJ Monthly,Minkerui

题意：给出N个数组，每个数组k个数，q个询问，要求判断两个数是否出现在同一个数组里面。

思路：对每个数字，用bitset在其出现过的数组都set为1，最后与一下两个数看是否非0即可。

# include <iostream># include <cstdio># include <bitset>using namespace std;bitset<1001>a[10001];int main(){    int n, k, t, q, x, y;    while(~scanf("%d",&n))    {        for(int i=0; i<=10000; ++i)            a[i].reset();        for(int i=0; i<n; ++i)        {            scanf("%d",&k);            while(k--)            {                scanf("%d",&t);                a[t].set(i);            }        }        scanf("%d",&q);        while(q--)        {            scanf("%d%d",&x,&y);            if((a[x]&a[y]).any())                puts("Yes");            else                puts("No");        }    }    return 0;}