1063 Set Similarity （25 分）求集合交集并集比值

版权声明：假装这里有个版权声明…… https://blog.csdn.net/CV_Jason/article/details/85329420

题目

Given two sets of integers, the similarity of the sets is defined to be $N_c/N_t\times 100$, 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%

解题思路

　　题目大意： 给定N个集合，求任意连个集合的交集元素比上并集元素个数的比值。
　　解题思路： 这道题的思路并不难，关键在于怎么实现集合的交集和并集，以及如何实现集合。直接使用stl中的set能够快速的解决这一问题。我们直接在此基础上统计不同的元素个数，即得到集合交集的元素个数，然后，我们直接用两个集合元素个数之和减去交集个数，得到并集个数，最后直接比值即可。

/*
** @Brief:No.1063 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-12-28
** @Solution: https://blog.csdn.net/CV_Jason/article/details/85329420
*/

#include<iostream>
#include<set>
#include<map>
using namespace std;

int main(){
	int N,M,K,temp;
	while(scanf("%d",&N)!=EOF){
		set<int> st[N];
		for(int i=0;i<N;i++){
			scanf("%d",&M);
			for(int j=0;j<M;j++){
				scanf("%d",&temp);
				st[i].insert(temp);
			}
		}
		int a,b;
		scanf("%d",&K);
		for(int i=0;i<K;i++){
			scanf("%d%d",&a,&b);
			int nc=0,nt=0;
			for(auto it = st[a-1].begin();it!=st[a-1].end();it++){
				if(st[b-1].find(*it)!=st[b-1].end()){
					nc++;
				}
			}
			nt = st[b-1].size()+st[a-1].size()-nc;
			
			printf("%.1f%%\n",nc*100.0/nt);
		}
		
	}
	return 0;
}

总结

　　最开始用set+map的组合进行集合运算，使用map是为了方便统计不同元素的个数，用元素出现次数作为value，结果最后一个点超时了，由于我对set和map的时空性能不是很熟悉，看了别人的答案，有set+find算法过的，看来map还是很耗时的。