問題 A: Set Similarity (25)

時間限制: 1 Sec  記憶體限制: 32 MB
提交: 252  解決: 106
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題目描述

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.

輸入

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.

輸出

For each query, print in one line the similarity of the sets, in the percentage form accurate up to 1 decimal place.

樣例輸入

3
3 99 87 101
4 87 101 5 87
7 99 101 18 5 135 18 99
2
1 2
1 3

樣例輸出

50.0%
33.3%

#include<iostream>
#include<set>
using namespace std;
//原想用map，沒想到直接set就過了
int main() {
	int n, k;
	while (cin >> n) {
		set<int> s[100];
		for (int i = 1; i <= n; i++) {
			int num;
			cin >> num;
			for (int j = 0; j < num; j++) {
				int x;
				cin >> x;
				s[i].insert(x);
			}
		}
		cin >> k;
		while (k--) {
			int x, y;
			cin >> x >> y;
			int Nc = 0, Nt = s[y].size();
			for (set<int>::iterator it = s[x].begin(); it != s[x].end(); it++) {
				if (s[y].find(*it) != s[y].end()) Nc++;
				else Nt++;
			}
			printf("%.1f%%\n", 100.0*Nc / Nt);
		}
	}
	return 0;
}