PAT甲级练习1063. Set Similarity (25)

最新推荐文章于 2021-08-10 19:55:39 发布

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最新推荐文章于 2021-08-10 19:55:39 发布

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本文介绍了一个计算两个整数集合相似度的问题，通过定义Nc为两个集合中共同拥有的不同整数数量，Nt为两个集合中不同整数的总数，从而计算出集合相似度。输入包括多个集合及其包含的元素，以及需要比较相似度的集合对。

1063. Set Similarity (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

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%

一开始想用合并set后计算大小与原先两个大小的关系来计算，但最后一个超时了，改用遍历和find的方法来做，没想到这样反而耗时要少一点。。

#include <iostream>  
#include <cstdio>  
#include <algorithm>  
#include <vector>  
#include <map>  
#include <set>  
#include <stack>  
#include <queue>  
#include <string>  
#include <string.h> 
using namespace std;

const int MAX = 1e5+10;

set<int> s[51];

//void query(int a, int b){
//	int a1 = s[a].size(), b1 = s[b].size();
//	int c1 = a1 + b1;
//	set<int> x(s[a]);
//	x.insert(s[b].begin(), s[b].end());
//	int c = x.size();
//	printf("%.1lf%%\n", (c1-c)*1.0/c*100);
//}

int main() {

	int n, m, d, k, k1, k2;
	char c[10];
	scanf("%d", &n);
	for(int i=1; i<=n; i++){
		scanf("%d", &m);
		for(int j=0; j<m; j++){
			scanf("%d", &d);
			s[i].insert(d);
		}	
	}
		
	scanf("%d", &k);
	for(int i=1; i<=k; i++){
		scanf("%d %d", &k1, &k2);
		int ns=0, nt=s[k2].size();
		for(auto it=s[k1].begin(); it!=s[k1].end(); it++){
			if(s[k2].find(*it)==s[k2].end())
				nt++;
			else
				ns++;
		}
		double ans = ns * 1.0 / nt *100;
		printf("%.1f%%\n", ans);
	}

	scanf("%d",&n);
    return 0;
}