【算法笔记】A1063 Set Similarity

 

1063 Set Similarity （25 分)

 

Given two sets of integers, the similarity of the sets is defined to be /, 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 (≤) which is the total number of sets. Then N lines follow, each gives a set with a positive M (≤) and followed by M integers in the range [0]. After the input of sets, a positive integer K (≤) 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%

题意：

　　比较两个set<int>集合的相似度。

code：

#include<bits/stdc++.h>
using namespace std;
vector<set<int> > sets;
int main(){
    int n,m,k,a1,a2;
    scanf("%d",&n);
    sets.resize(n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d",&m);
        set<int> temp;
        for (int j=0;j<m;j++)
        {
            int num;
            scanf("%d",&num);
            temp.insert(num);
        }
        sets[i]=temp;
    }
    scanf("%d",&k);
    for (int i=0;i<k;i++)
    {
        scanf("%d %d",&a1,&a2);
        int nc=0,nt = sets[a2-1].size();
        for (auto it=sets[a1-1].begin();it!=sets[a1-1].end();it++)
        {
            if (sets[a2-1].find(*it)==sets[a2-1].end())
                nt++;
            else
                nc++;
        }
        printf("%.1lf%%\n",(double)nc/nt*100);
    }
    return 0;
}