POJ2443 Set Operation —— bitset

题目链接：https://vjudge.net/problem/POJ-2443

Set Operation

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3554		Accepted: 1477

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

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个集合，每个集合有若干个数。有m个询问，问x、y是否存在于同一个集合中。

题解：

C++ bitset的应用。

具体介绍：https://blog.csdn.net/qll125596718/article/details/6901935

成员函数	函数功能
bs.any()	是否存在值为1的二进制位
bs.none()	是否不存在值为1的二进制位 或者说是否全部位为0
bs.size()	位长，也即是非模板参数值
bs.count()	值为1的个数
bs.test(pos)	测试pos处的二进制位是否为1 与0做或运算
bs.set()	全部位置1
bs.set(pos)	pos位处的二进制位置1 与1做或运算
bs.reset()	全部位置0
bs.reset(pos)	pos位处的二进制位置0 与0做或运算
bs.flip()	全部位逐位取反
bs.flip(pos)	pos处的二进制位取反
bs.to_ulong()	将二进制转换为unsigned long输出
bs.to_string()	将二进制转换为字符串输出
~bs	按位取反 效果等效为bs.flip()
os << b	将二进制位输出到os流 小值在右，大值在左

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #include <bitset>   //bitset头文件
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 1e5+;

 bitset<>a[];
 int main()
 {
     int n;
     while(scanf("%d", &n)!=EOF)
     {
         for(int i = ; i<; i++)
             a[i].reset();
         for(int i = ; i<=n; i++)
         {
             int m, x;
             scanf("%d", &m);
             while(m--)
             {
                 scanf("%d", &x);
                 a[x][i] = ;
             }
         }

         int m, x, y;
         scanf("%d", &m);
         while(m--)
         {
             scanf("%d%d", &x,&y);
             if((a[x]&a[y]).count()) puts("Yes");
             else puts("No");
         }
     }
 }