poj2443Set Operation (bitset)

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

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

题意：有n个集合，每个集合里有c[i]个数，可能重复，共有m个操作，每个操作询问两个数，问这两个数是否在n个集合中的某一个同时出现。

思路：一开始标记每一个集合中出现的数，然后O(n*m)的复杂度T了，换了bitset的思路，即用bitset<1005>bt[10005]记录第i个元素在第j个集合出现的情况，然后对于任意两个数a,b，只要用(bt[a]&bt[b]).any()判断一下是否出现过就行。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
#include<bitset>
#define inf 99999999
#define pi acos(-1.0)
#define maxn 1005
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef long double ldb;
bitset<1005>bt[10005];

int main()
{
    int n,m,i,j,c,d;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=10000;i++)bt[i].reset();
        for(i=1;i<=n;i++){
            scanf("%d",&c);
            for(j=1;j<=c;j++){
                scanf("%d",&d);
                bt[d][i]=1;
            }
        }
        scanf("%d",&m);
        for(i=1;i<=m;i++){
            scanf("%d%d",&c,&d);
            if((bt[c]&bt[d]).any() )printf("Yes
");
            else printf("No
");

        }
    }
    return 0;
}