PAT (Advanced Level) Practise 1063 Set Similarity (25)

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本文介绍了一个计算两个整数集合相似度的问题，通过定义Nc/Nt*100%来衡量两个集合的相似程度，其中Nc是共同元素数量，Nt是总的不同元素数量。文章提供了使用C++ set进行高效求解的方法。

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1063. Set Similarity (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<=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%

题意：给定两个整数集合，它们的相似度定义为：N_c/N_t*100%。其中N_c是两个集合都有的不相等整数的个数，N_t是两个集合一共有的不相等整数的个数。就是计算任意一对给定集合的相似度

解题思路：set的应用

#include <iostream>  
#include <cstdio>  
#include <string>  
#include <cstring>  
#include <algorithm>  
#include <cmath>  
#include <queue>  
#include <vector>  
#include <set>  
#include <stack>  
#include <map>  
#include <climits>  
  
using namespace std;  
  
#define LL long long  
  
const int INF=0x3f3f3f3f;  
  
int main()  
{  
    int n;  
    while(~scanf("%d",&n))  
    {  
        int k,a;  
        set<int>s[60];  
        for(int i=1;i<=n;i++)  
        {  
            scanf("%d",&k);  
            for(int j=1;j<=k;j++)  
                scanf("%d",&a),s[i].insert(a);  
        }  
        int m;  
        scanf("%d",&m);  
        while(m--)  
        {  
            int x,y;  
            scanf("%d %d",&x,&y);  
            set<int>::iterator it;  
            int sum=0;  
            for(it=s[x].begin();it!=s[x].end();it++)  
                if(s[y].count(*it)) sum++;  
            printf("%.1lf%%\n",1.0*sum/(s[x].size()+s[y].size()-sum)*100);  
        }  
    }  
    return 0;  
}