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[];
 int Size[];
 int main()
 {
     int N;
     cin >> N;
     for (int i = ; i < N; i++)
     {
         int K;
         cin >> K;
         for (int j = ; j <K; j++)
         {
             int num;
             cin >> num;
             M[i][num]++;
             Size[i]++;
         }
     }
     int K;
     cin >> K;
     for (int i = ; i < K; i++)
     {
         int v1, v2;
         int trueHave = ;  //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%%\n", (1.0 * trueHave) / (1.0 * (size-trueHave))*100.0);
     }
 }