pat1063. Set Similarity (25)

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%

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提交代码

 

#include<cstdio>
#include<stack>
#include<cstring>
#include<iostream>
#include<stack>
#include<set>
#include<map>
using namespace std;
map<int,set<int> > ha;
int main(){
    //freopen("D:\\INPUT.txt","r",stdin);
    int n,m;
    scanf("%d",&n);
    int i,j,num;
    for(i=1;i<=n;i++){
        scanf("%d",&m);
        for(j=0;j<m;j++){
            scanf("%d",&num);
            ha[i].insert(num);
        }
    }
    /*for(i=1;i<=n;i++){
        cout<<"i:  "<<i<<endl;
        set<int>::iterator it;
        for(it=ha[i].begin();it!=ha[i].end();it++){
            cout<<*it<<endl;
        }
    }*/
    scanf("%d",&m);
    int a,b;
    set<int>::iterator it;
    for(i=0;i<m;i++){
        scanf("%d %d",&a,&b);
        int sum=0;
        for(it=ha[a].begin();it!=ha[a].end();it++){
            if(ha[b].count(*it)){
                sum++;
            }
        }
        //cout<<sum<<" "<<ha[b].size()+ha[a].size()<<endl;
        printf("%.1lf%\n",sum*1.0*100/(ha[b].size()+ha[a].size()-sum));
    }
    return 0;
}