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.
题意：求X，Y是否在同一集合出现过，注意X=Y时，出现过一次就满足了。
思路：用bitset来表示元素在哪些集合出现过，如果X和Y出现的集合有交集，则满足。
积累一下：
s.set()是全部置为1，s.set(pos)是某一位置为1，s.reset()是全部置为0.
s.flip()，按位取反。
#include<cstdio> #include<bitset> #include<cstdlib> #include<iostream> #include<algorithm> using namespace std; const int maxn=10000; bitset<1010>s[maxn+10]; int read() { char c=getchar(); int res; while(c>‘9‘||c<‘0‘) c=getchar(); for(res=0;c>=‘0‘&&c<=‘9‘;c=getchar()) res=(res<<3)+(res<<1)+c-‘0‘; return res; } int main() { int N,Q,i,num,x,y; while(~scanf("%d",&N)){ for(i=1;i<=maxn;i++) s[i].reset(); for(i=1;i<=N;i++){ num=read(); while(num--){ x=read(); s[x][i]=1; //s[x].set(i); } } Q=read(); while(Q--){ x=read(); y=read(); if((s[x]&s[y]).count()) puts("Yes"); //.any() else puts("No"); } } return 0; }
