pat advanced level 1063
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1063 Set Similarity (25)(25 分)
Given two sets of integers, the similarity of the sets is defined to be N~c~/N~t~*100%, where N~c~ is the number of distinct common numbers shared by the two sets, and N~t~ is the total number of distinct numbers in the two sets. Your job is to calculate the similarity of any given pair of sets.
Input Specification:
Each input file contains one test case. Each case first gives a positive integer N (<=50) which is the total number of sets. Then N lines follow, each gives a set with a positive M (<=10^4^) and followed by M integers in the range [0, 10^9^]. After the input of sets, a positive integer K (<=2000) is given, followed by K lines of queries. Each query gives a pair of set numbers (the sets are numbered from 1 to N). All the numbers in a line are separated by a space.
Output Specification:
For each query, print in one line the similarity of the sets, in the percentage form accurate up to 1 decimal place.
Sample Input:
3
3 99 87 101
4 87 101 5 87
7 99 101 18 5 135 18 99
2
1 2
1 3
Sample Output:
50.0%
33.3%
/********************** author: yomi date: 18.8.2 ps: 使用algorithm下的count函数会超时 得用set下的count **********************/ #include <iostream> #include <set> #include <algorithm> #include <cstdio> using namespace std; set<int>s[60]; int main() { int n, k, m, t; scanf("%d", &n); for(int i=0; i<n; i++){ scanf("%d", &m); for(int j=0; j<m; j++){ scanf("%d", &t); s[i].insert(t); } } scanf("%d", &k); int a, b; for(int i=0; i<k; i++){ scanf("%d%d", &a, &b); int c = 0, d = 0; set<int>::iterator iter; for(iter=s[a-1].begin(); iter!=s[a-1].end(); ++iter){ if(s[b-1].count(*iter)){ c++; } } int nc = c; int nt = s[a-1].size()+s[b-1].size()-c; double ans = nc*1.0/nt*100; printf("%.1f%% ", ans); } return 0; } /** 3 3 99 87 101 4 87 101 5 87 7 99 101 18 5 135 18 99 2 1 2 1 3 **/
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