1063 Set Similarity (25分)
Given two sets of integers, the similarity of the sets is defined to be Nc/Nt×100%, where Nc is the number of distinct common numbers shared by the two sets, and Nt 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 (≤104) and followed by M integers in the range [0,109]. 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%
题目大意:
给定N个集合,给出的集合中可能含有相同的值。然后要求M个查询,每个查询给出两个集合的编号x和y,要求集合x和集合y的相同元素率:。
解题思路:
这里我们可以借助N个STL的set容器来保存这N个set,当判断x和y两个set的相同元素率时,我们只需要遍历这两个set统计其中的重复元素的个数即可。
#include<iostream>
#include<cstdio>
#include<set>
using namespace std;
set<int> s[60];
double fun(int x, int y)
{
int cnt = 0;
for(set<int>::iterator it = s[x].begin();it!=s[x].end();it++)
{
if(s[y].find(*it)!=s[y].end())
cnt++;
}
double res = cnt*100.0 / (s[x].size() + s[y].size() - cnt);
return res;
}
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
int m;
scanf("%d", &m);
for (int j = 0; j < m; j++)
{
int num;
scanf("%d", &num);
s[i].insert(num);
}
}
int k;
scanf("%d", &k);
for (int i = 0; i < k; i++)
{
int x, y;
scanf("%d %d", &x, &y);
printf("%.1lf%%\n", fun(x, y));
}
return 0;
}