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%%\n", 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
**/
原文地址：https://www.cnblogs.com/AbsolutelyPerfect/p/9409417.html