1063 Set Similarity——PAT甲级

1063 Set Similarity

Given two sets of integers, the similarity of the sets is defined to be Nc/Nt*100%, where Nc is the number of distinct common numbers shared by the two sets, and Nt 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%

题目大意：定义集合的相似度为Nc / Nt * 100%,其中Nc为两个集合相同数字的个数，Nt为两个集合不同数字的总数。给出N个集合，然后有K次查询，让你求出所查两个集合的相似度。

大致思路：先利用set存储每个集合，然后执行K次查询是利用find函数来查找集合相同元素。

代码：

#include <bits/stdc++.h>

using namespace std;

int n,k;

int main() {
    scanf("%d", &n);
    set<int> se[n + 1];
    for (int i = 0; i < n; i++) {
        int m; scanf("%d", &m);
        for (int j = 0; j < m; j++) {
            int x; scanf("%d", &x);
            se[i + 1].insert(x);
        }
    }
    scanf("%d",&k);
    for (int i = 0; i < k; i++) {
        int q1, q2; scanf("%d%d",&q1, &q2);
        double Nc = 0.0, Nt = 0.0;
        for (set<int> :: iterator ite = se[q1].begin() ; ite != se[q1].end(); ite++) {
            if (se[q2].find(*ite) != se[q2].end()) Nc++;
        }
        Nt = (se[q1].size() + se[q2].size()) * 1.0 - Nc;
        double ans = Nc / Nt * 100.0;
        printf("%.1lf",ans);puts("%");
    }
    return 0;
}