PAT-A-1063 Set Similarity 【set】

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%

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看到distinct就想到了set的使用。

我当时这样想，使用set数组保存每个数字序列，当让求某两个序列的相似度时，使用两个set变量，one、two分别等于这两个序列，然后将一个序列插入到一个另一个序列中，这样分母就是得到的新序列的长度，分子就是 one、two插入操作之前的size之和减去插入操作之后的size，但是很遗憾，由于多次使用set进行操作，有一组测试用例超时了。

#include 
#include 
using namespace std;
int main(){
	int num,temp,a;
	set res[50];
	scanf("%d",&num);
	for(int i=0;i temp_one,temp_two;
	scanf("%d",&m);
	while(m--){
		scanf("%d %d",&one,&two);
		temp_one=res[one-1];
		temp_two=res[two-1];
		int sizea=temp_one.size();
		int sizet=temp_two.size();
		temp_one.insert(temp_two.begin(),temp_two.end());
		int size=temp_one.size();
		printf("%.1lf%\n",100*((double)sizea+sizet-temp_one.size())/temp_one.size());
	}
	return 0;
}

既然找到了引起超时的原因是因为大量的使用set运算，那么针对于这部分进行改进，使用遍历的思路。 

下面是查询过程，使用指针遍历set，使用count查询是否为公共元素

int m,count=0;
	scanf("%d",&m);
	while(m--){
		count=0;
		scanf("%d %d",&one,&two);
		for(set::iterator lt=res[one-1].begin();lt!=res[one-1].end();lt++){
			if(res[two-1].count(*lt)==1) count++;
		}
		printf("%.1lf%%\n",100*((double)count)/(res[one-1].size()+res[two-1].size()-count));
	}