1063 Set Similarity (25point(s))
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%
超时
#include<bits/stdc++.h>
using namespace std;
set<int> st[55];
int main(){
int n,m;
cin>>n;
for(int i=1;i<=n;++i){
int num,k;
cin>>k;
while(k--){
cin>>num;
st[i].insert(num);
}
}
cin>>m;
int first,second;
for(int i=0;i<m;++i){
set<int> temp;
cin>>first>>second;
for(set<int>::iterator it=st[first].begin();it!=st[first].end();++it)
temp.insert(*it);
for(set<int>::iterator it=st[second].begin();it!=st[second].end();++it)
temp.insert(*it);
printf("%.1lf%\n",(st[first].size()+st[second].size()-temp.size())*100.0/temp.size());
}
}
#include<bits/stdc++.h>
using namespace std;
set<int> st[55];
int main(){
int n,m;
cin>>n;
for(int i=1;i<=n;++i){
int num,k;
cin>>k;
while(k--){
cin>>num;
st[i].insert(num);
}
}
cin>>m;
int first,second;
for(int i=0;i<m;++i){
int sameNum=0;
cin>>first>>second;
for(auto it=st[first].begin();it!=st[first].end();++it){
if(st[second].find(*it)!=st[second].end()) sameNum++;
}
printf("%.1lf%\n",sameNum*100.0/(st[first].size()+st[second].size()-sameNum));
}
}