PAT A 1063. Set Similarity (25)

题目

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%

 

即统计两个数串不重复的相同数字占总数字（不重复）的比重。

 

代码：

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

int main()
{
	set<int> sets[50];	//输入的sets，利用set直接排好序，且唯一了
	
	int n,m,i,j,temp;	//输入数据
	cin>>n;
	for(i=0;i<n;i++)
	{
		scanf("%d",&m);
		for(j=0;j<m;j++)
		{
			scanf("%d",&temp);
			sets[i].insert(temp);
		}
	}

	set<int>::iterator p1,p2;	//迭代器，用于扫描
	int k;
	int s1,s2;
	int nc,nt;	//nc,nt
	double rate;
	cin>>k;
	for(i=0;i<k;i++)
	{
		scanf("%d %d",&s1,&s2);
		p1=sets[s1-1].begin();	//从头开始扫描
		p2=sets[s2-1].begin();
		nc=0;
		nt=0;
		while(p1!=sets[s1-1].end()&&p2!=sets[s2-1].end())	//扫描到一个结束
		{
			if(*p1==*p2)	//相等
			{
				nc++;
				nt++;
				p1++;
				p2++;
			}
			else if(*p1<*p2)	//两种不等
			{
				nt++;
				p1++;
			}
			else
			{
				nt++;
				p2++;
			}
		}
		while(p1!=sets[s1-1].end())	//处理没结束的一个的剩余元素数量
		{
			p1++;
			nt++;
		}
		while(p2!=sets[s2-1].end())
		{
			p2++;
			nt++;
		}
		rate=double(nc)*100/nt;	//计算、输出
		printf("%0.1f%%\n",rate);
	}

	return 0;
}