PAT A1063

1063. Set Similarity (25)

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:

33 99 87 1014 87 101 5 877 99 101 18 5 135 18 9921 21 3

Sample Output:

50.0%33.3%

实现代码如下：

#include <cstdio>#include <set>using namespace std;set<int> gz[55];void show(int a, int b){    int total = gz[b].size(), num = 0;    for(set<int>::iterator it = gz[a].begin(); it != gz[a].end(); it++){        if(gz[b].find(*it) != gz[b].end()) num++;        else total++;    }    printf("%.1f%%\n", num * 100.0 / total);}int main(){    int N, M, temp;    scanf("%d", &N);    for(int i = 1; i <= N; i++){        scanf("%d", &M);        for(int j = 0; j < M; j++){            scanf("%d", &temp);            gz[i].insert(temp);        }    }    int K, a, b;    scanf("%d", &K);    for(int i = 0; i < K; i++){        scanf("%d%d", &a, &b);        show(a, b);    }    return 0;}