1063 Set Similarity （25 分）
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 (≤10?4?? ) and followed by M integers in the range [0,10?9?? ]. 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%
題目大意：
本題定義集合相似度Nc/Nt*100%,其中Nc是兩個集合中共有的不相等元素個數(shù)旗吁，Nt是兩個集合中不相等元素的個數(shù)废累。你的任務(wù)是計算給定的集合的相似度十饥。
#include <iostream>
#include <vector>
#include <set>
using namespace std;
int main(){
int n;
cin>>n;
vector<set<int> > v(n);
for(int i=0;i<n;i++){
int m;
cin>>m;
while(m--){
int val;
cin>>val;
v[i].insert(val);
}
}
int k;
cin>>k;
for(int i=0;i<k;i++){
int a,b;
cin>>a>>b;
a--,b--;
int cnt=0;
for(auto it:v[a]){
if(v[b].find(it)!=v[b].end()) cnt++;
}
printf("%.1lf%\n",cnt*100.0/(v[a].size()+v[b].size()-cnt));
}
return 0;
}
