poj 2443
Set Operation
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 2961 | Accepted: 1192 |
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.
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<algorithm> #include<string> #include<vector> using namespace std; int n,m; int a[32][10005]; int main() { int num,tt,q; while(scanf("%d",&n)!=EOF) { bool flag; memset(a,0,sizeof(a)); for(int i=0;i<n;i++) { scanf("%d",&num); int xx,yy; xx=i/32; yy=i%32; for(int j=0;j<num;j++) { scanf("%d",&tt); a[xx][tt]|=(1<<yy); } } m=n/32; scanf("%d",&q); for(int i=0;i<q;i++) { int u,v; flag=false; scanf("%d%d",&u,&v); for(int j=0;j<=m;j++) { if(a[j][u]&a[j][v]) { flag=1; break; } } if(flag) printf("Yes\n"); else printf("No\n"); } } return 0; }