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%
题意:给你N组数据,共有K次询问,每次询问给出一对数 x,y 意思是在数组x与数组y中求出 Nc/Nt×100% ,Nc表示数组x,y中元素 的交集,Nt表示数组x,y中元素的并集
思路:STL 中set容器有三种函数能满足题目要求
1.set_union(取两集合并集)
2.set_intersection(取两集合交集)
3.set_difference(取两集合差集)
代码
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<set>
using namespace std;
const int Max=1e4+5;
int main()
{
int N,M,Q,x,y,Nc,Nt;
scanf("%d",&N);
set<int> *S=new set<int>[N+1];
for(int i=1;i<=N;i++)
{
scanf("%d",&M);
while(M--)
{
scanf("%d",&x);
S[i].insert(x);
}
}
scanf("%d",&Q);
set<int> ans;
while(Q--)
{
scanf("%d %d",&x,&y);
set_intersection(S[x].begin(),S[x].end(),S[y].begin(),S[y].end(),inserter(ans,ans.begin()));
Nc=ans.size();
Nt=S[x].size()+S[y].size()-Nc;
ans.clear();
printf("%.1lf%%\n",100*(1.0*Nc/Nt));
}
return 0;
}