1063 Set Similarity (25分)

Given two sets of integers, the similarity of the sets is defined to be /, 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 (≤) which is the total number of sets. Then N lines follow, each gives a set with a positive M (≤) and followed by M integers in the range [0]. After the input of sets, a positive integer K (≤) 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%

题目分析：利用map对应存储需要的键与值 然后从键少的那个遍历 看另一个map中有没有对应的值 注
不能通过直接访问来判断某个键值是否存在(在map<int int>的情况下 访问不存在的键 会生成相应的键值对 其值默认为0）要通过map中find函数来判断键值是否存在

#define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
map<int, int> M[50];
int Size[50];
int main()
{
    int N;
    cin >> N;
    for (int i = 0; i < N; i++)
    {
        int K;
        cin >> K;
        for (int j = 0; j <K; j++)
        {
            int num;
            cin >> num;
            M[i][num]++;
            Size[i]++;
        }
    }
    int K;
    cin >> K;
    for (int i = 0; i < K; i++)
    {
        int v1, v2;
        int trueHave = 0;  //相等元素的个数
        cin >> v1 >> v2;    
        v1--;
        v2--;
        int size = M[v1].size() + M[v2].size();
        int v = (Size[v1] < Size[v2]) ? v1 : v2;
        int d = (Size[v1] < Size[v2]) ? v2 : v1;
        for (auto it : M[v])
            if (M[d].find(it.first)!=M[d].end())
                trueHave++;
        printf("%.1f%%
", (1.0 * trueHave) / (1.0 * (size-trueHave))*100.0);
    }
}