Set Operation
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 3558 | Accepted: 1479 |
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
有n个集合，给出两个数，判断这两个数在不在同一个集合中。
分析
使用bitset来进行交集操作。b.any()为b中是否存在置为1的二进制位？注意，两个bitset求交集后返回的还是一个bitset。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
#include<map>
#include<deque>
using namespace std;
typedef long long LL;
const int maxn = 1e4+;
const int mod = +;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
//#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))
const int inf = 0x3f3f3f3f;
#define lson l,m,2*rt
#define rson m+1,r,2*rt+1
//每个bitset的大小为1024
bitset<> b[maxn];
int main(){
int n,x,y,q;
while(~scanf("%d",&n)){
for(int i=;i<maxn;i++) b[i].reset(); for(int i=;i<n;i++){
scanf("%d",&q);
while(q--){
scanf("%d",&x);
b[x][i]=;
} }
scanf("%d",&q);
while(q--){
scanf("%d%d",&x,&y); if((b[x]&b[y]).any()){
puts("Yes");
}else {
puts("No");
} }
}
return ;
}