【PAT】1063. Set Similarity (25) 待改进

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%

分析：两个集合共有的元素个数 / 两个集合合并后的元素个数.

有一组超时，待改进：

#include<iostream>

#include<set>

#include<vector>

using namespace std;

int main()

{

	int N;

	int i,count,j,t;

	cin>>N;

	vector< set<int> > vec(N);

	for(i=0; i<N; i++){

		scanf("%d",&count);

		for(j=0; j<count; j++){

			scanf("%d",&t);

			vec[i].insert(t);

		}

	}

	int request,a,b;

	scanf("%d",&request);

	for(i=0; i<request; i++){

		scanf("%d %d",&a,&b);

		set<int> temp_set;

		set<int>::iterator it;

		for(it = vec[a-1].begin(); it != vec[a-1].end(); it++){

			temp_set.insert( *it );

		}

		for(it = vec[b-1].begin(); it != vec[b-1].end(); it++){

			temp_set.insert( *it );

		}

		printf( "%.1f",(float)( vec[a-1].size()+vec[b-1].size()-temp_set.size())*100/(float)(temp_set.size()) );

		cout<<"%"<<endl;

	}

	return 0;

}

巴特西

【PAT】1063. Set Similarity (25) 待改进

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