Pat(Advanced Level)Practice--1063(Set Similarity)

Pat1063代码

题目描述：

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%

利用map和set的性质进行筛选，结果最后一个case超时。

代码：

#include<cstdio>
#include<map>
#include<set>
#define N 55

using namespace std;

set<int> s[N];

int main(int argc,char *argv[])
{
	int n;
	int i,j;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		int m;
		int temp;
		scanf("%d",&m);
		for(j=1;j<=m;j++)
		{
			scanf("%d",&temp);
			s[i].insert(temp);
		}
	}
	int k;
	scanf("%d",&k);
	while(k--)
	{
		map<int,int> m;
		set<int>::iterator p;
		int x,y;
		scanf("%d%d",&x,&y);
		for(p=s[x].begin();p!=s[x].end();p++)
		{
			int index=*p;
			m[index]++;
		}
		for(p=s[y].begin();p!=s[y].end();p++)
		{
			int index=*p;
			m[index]++;
		}
		int count=0;
		map<int,int>::iterator it;
		for(it=m.begin();it!=m.end();it++)
			if(it->second>1)
				count++;
		printf("%.1f%%\n",100*count*1.0/m.size());
	}

	return 0;
}

AC代码：

利用函数求两个集合的交集，然后就可以求出集合的相似性了；

#include<cstdio>
#include<set>
#include<iterator>
#include<algorithm>
#define N 55

using namespace std;

set<int> s[N];

int main(int argc,char *argv[])
{
	int n;
	int i,j;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		int m;
		int temp;
		scanf("%d",&m);
		for(j=1;j<=m;j++)
		{
			scanf("%d",&temp);
			s[i].insert(temp);
		}
	}
	int k;
	scanf("%d",&k);
	while(k--)
	{
		int x,y;
		int ret,total;
		set<int> ans;
		scanf("%d%d",&x,&y);
		set_intersection(s[x].begin(),s[x].end(),s[y].begin(),s[y].end(),
		inserter(ans,ans.begin()));//求集合交集
		ret=ans.size();//交集元素个数
		total=s[x].size()+s[y].size()-ret;//不重复元素总个数
		printf("%.1f%%\n",100*ret*1.0/total);
	}

	return 0;
}

最后一个case跑了大概130ms;

AC代码：

写一个函数求集合的并集，其实这和上面的方法是一样的，只不过一个是交集，一个是并集；

#include<cstdio>
#include<set>
#include<iterator>
#include<algorithm>
#define N 55

using namespace std;

set<int> s[N];

int GetUnion(int x,int y)//求并集元素的个数，其实algorithm里面有set_union
{                        //这个函数，在这里我们可以简单的实现一下
	int ret=0;
	set<int>::iterator it;
	for(it=s[x].begin();it!=s[x].end();it++)
	{
		if(s[y].find(*it)==s[y].end())
			ret++;
	}
	ret+=s[y].size();
	return ret;
}

int main(int argc,char *argv[])
{
	int n;
	int i,j;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		int m;
		int temp;
		scanf("%d",&m);
		for(j=1;j<=m;j++)
		{
			scanf("%d",&temp);
			s[i].insert(temp);
		}
	}
	int k;
	scanf("%d",&k);
	while(k--)
	{
		int x,y;
		int ret,total;
		scanf("%d%d",&x,&y);
		total=GetUnion(x,y);
		ret=s[x].size()+s[y].size()-total;
		printf("%.1f%%\n",100*ret*1.0/total);
	}
	return 0;
}

最后一个case跑了大概170ms;