PAT (Advanced Level) Practice 1063 Set Similarity（25分）【set、查询】

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 $M (≤10^4)$ and followed by $M$ integers in the range $[0,10^9]$. 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%

题意

求两个集合的交集与并集的大小。

思路

使用set就可以。

代码

#include <iomanip>
#include <iostream>
#include <set>

using namespace std;

set<int> sets[55];

int main() {
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        int t;
        cin >> t;
        for (int j = 0; j < t; ++j) {
            int num;
            cin >> num;
            sets[i].insert(num);
        }
    }

    cin >> n;
    for (int i = 0; i < n; ++i) {
        set<int> comm;
        int nc = 0, nt = 0, a, b;

        cin >> a >> b;
        for (auto &e : sets[a]) {
            if (sets[b].count(e))
                ++nc;
            else
                ++nt;
        }
        nt += sets[b].size();

        cout << fixed << setprecision(1) << 100.0 * nc / nt << "%\n";
    }
}