PAT1063:Set Similarity

Posted 2020-10-09 0kk470

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%

思路

求交集大小/并集大小的百分比，保留一位小数。

暂时直接用stl的set做，然而最后一个测试用例超时。。。
代码
待优化

#include<iostream>
#include<algorithm>
#include<iterator>
#include<iomanip>
#include<set>
#include<vector>
using namespace std;
int main()
{
    int N;
    while(cin >> N)
    {
        //input data
       vector<set<int>> sets(N + 1);
       for(int i = 1;i <= N;i++)
       {
           int M;
           cin >> M;
           for(int j = 0;j < M;j++)
           {
               int value;
               cin >> value;
               sets[i].insert(value);
           }
       }
       //output
       int K,first,second;

       cin >> K;
       for(int j = 0;j < K;j++)
       {
           cin >> first;
           cin >> second;
           set<int> u;
           set<int> i;

           set_union(sets[first].begin(),sets[first].end(),sets[second].begin(),sets[second].end(),inserter(u,u.begin())); //union
           set_intersection(sets[first].begin(),sets[first].end(),sets[second].begin(),sets[second].end(),inserter(i,i.begin())); //intersection

           double t = u.size();
           double c = i.size();
           cout << fixed << setprecision(1) << c/t*100 << "%" << endl;
       }
    }
}