【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 = ; i < n; i++)
{
scanf("%d",&m);
set<int> temp;
for (int j=;j<m;j++)
{
int num;
scanf("%d",&num);
temp.insert(num);
}
sets[i]=temp;
}
scanf("%d",&k);
for (int i=;i<k;i++)
{
scanf("%d %d",&a1,&a2);
int nc=,nt = sets[a2-].size();
for (auto it=sets[a1-].begin();it!=sets[a1-].end();it++)
{
if (sets[a2-].find(*it)==sets[a2-].end())
nt++;
else
nc++;
}
printf("%.1lf%%\n",(double)nc/nt*);
}
return ;
}
上一篇:[Hadoop] - 自定义Mapreduce InputFormat&OutputFormat


下一篇:SpringBoot 如何从前台传递数组