Set Operation
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 3606 | Accepted: 1499 |
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.
解析:起先想到并查集,但是并查集是把元素都并成一棵树,然后寻找根节点,这里却问的是在不在同一个集合。
可以想到,开一个数组,每30 个元素并在一个集合里。(状态压缩)我用到的是bitset维护每一个元素。bitset相关函数网上自寻。
代码:
#include<cstdio> #include<bitset> using namespace std; bitset<1005> bit[10005]; int main() { int n,m,x,y,q; while(~scanf("%d",&n)) { for(int i=0; i<10005; i++) bit[i].reset(); for(int i=0; i<n; i++) { scanf("%d",&m); for(int j=0; j<m; j++) { scanf("%d",&x); bit[x][i]=1; } } scanf("%d",&q); while(q--) { scanf("%d%d",&x,&y); if((bit[x]&bit[y]).count()!=0) { puts("Yes"); } else puts("No"); } } return 0; }