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最新推荐文章于 2020-03-27 15:07:14 发布

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本文介绍了一个关于猴子与香蕉的趣味算法问题。通过两份清单——猴子偏好香蕉类型及不同地点提供的香蕉类型，该算法旨在找出每只猴子是否能在某个地点找到其偏好的香蕉种类。文章提供了一段C++代码实现，包括输入输出格式和具体解决方法。

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Bananas are the favoured food of monkeys.

In the forest, there is a Banana Company that provides bananas from different places.

The company has two lists.

The first list records the types of bananas preferred by different monkeys, and the second one records the types of bananas from different places.

Now, the supplier wants to know, whether a monkey can accept at least one type of bananas from a place.

Remenber that, there could be more than one types of bananas from a place, and there also could be more than one types of bananas of a monkey's preference.

Input Format

The first line contains an integer $T$ , indicating that there are $T$ test cases.

For each test case, the first line contains two integers $N$ and $M$ , representing the length of the first and the second lists respectively.

In the each line of following $N$ lines, two positive integers $i, j$ indicate that the $i$ -th monkey favours the $j$ -th type of banana.

In the each line of following $M$ lines, two positive integers $j, k$ indicate that the $j$ -th type of banana could be find in the $k$ -th place.

All integers of the input are less than or equal to $50$ .

Output Format

For each test case, output all the pairs $x, y$ that the $x$ -the monkey can accept at least one type of bananas from the $y$ -th place.

These pairs should be outputted as ascending order. That is say that a pair of $x, y$ which owns a smaller $x$ should be output first.

If two pairs own the same $x$ , output the one who has a smaller $y$ first.

And there should be an empty line after each test case.

样例输入

样例输出

#include<iostream>
#include<algorithm>
using namespace std;
typedef struct mtb
{
	int mnum;
	int bnum;
}mtb;
mtb a[55];
typedef struct btr{
	int bnum;
	int rnum;
}btr;
btr b[55];
typedef struct M{
	int mnum;
	int rnum;
}M;
M F[1100];
int cmp(M p1, M p2)
{
    if (p1.mnum < p2.mnum)
    {
        return 1;
    }
    if (p1.mnum == p2.mnum && p1.rnum < p2.rnum)
    {
        return 1;
    }
    return 0;
}
int  x=0;
int main()
{
	int T;
	int list1,list2;
	cin>>T;
	while(T--)
	{
		cin>>list1>>list2;
		for(int i=0;i<list1;i++)
		{
			cin>>a[i].mnum;
			cin>>a[i].bnum;
		}
		for(int j=0;j<list2;j++)
		{
			cin>>b[j].bnum;
			cin>>b[j].rnum;
		} 	
		 for(int k=0;k<list1;k++)
		 {
		        for(int q=0;q<list2;q++)
				{
					if(a[k].bnum == b[q].bnum)
					{
						F[x].mnum=a[k].mnum;
						F[x].rnum=b[q].rnum;
						x++;
					}		
				} 		
		 }	
		 sort(F,F+x,cmp); 
		 for(int t=1;t<=x;t++){
		 	    if(!(F[t].mnum==F[t-1].mnum && F[t].rnum==F[t-1].rnum))
		 		cout<<F[t-1].mnum<<" "<<F[t-1].rnum<<endl;
		 }
		 cout<<endl;
		 x=0;
	}
}