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");
}
}
}

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