POJ2443 Set Operation （基础bitset应用，求交集）

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

3
3 1 2 3
3 1 2 5
1 10
4
1 3
1 5
3 5
1 10

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.

题意：给定N个集合，Q次询问，对每次询问，求X，Y是否在同一集合出现过，注意X=Y时，X在一个集合里至少出现一次就满足了。

思路：用Bitset来表示每个元素在哪些集合出现过，如果X和Y出现的集合有交集，则满足。

积累一下：

s.set()是全部置为1，s.set(pos)是某一位置为1，s.reset()是全部置为0.

s.flip()，按位取反。

#include<cstdio>
#include<bitset>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
bitset<>s[maxn+];
int read()
{
    char c=getchar(); int res;
    while(c>''||c<'') c=getchar();
    for(res=;c>=''&&c<='';c=getchar()) res=(res<<)+(res<<)+c-'';
    return res;
}
int main()
{
    int N,Q,i,num,x,y;
    while(~scanf("%d",&N)){
        for(i=;i<=maxn;i++) s[i].reset();
        for(i=;i<=N;i++){
            num=read();
            while(num--){
                x=read();
                s[x][i]=; //s[x].set(i);
            }
        }
        Q=read();
        while(Q--){
            x=read(); y=read();
            if((s[x]&s[y]).count()) puts("Yes"); //.any()
            else puts("No");
        }
    }
    return ;
}