2017 icpc 乌鲁木齐赛区 A.Banana （签到题）

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概述

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, ji,j indicate that the ii-th monkey favours the j-th type of banana.
In the each line of following M lines, two positive integers j, kj,k indicate that the jj-th type of banana could be find in the kk-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, yx,y that the xx-the monkey can accept at least one type of bananas from the yy-th place.
These pairs should be outputted as ascending order. That is say that a pair of x, yx,y which owns a smaller xx should be output first.
If two pairs own the same xx, output the one who has a smaller yy first.
And there should be an empty line after each test case.
样例输入
1
6 4
1 1
1 2
2 1
2 3
3 3
4 1
1 1
1 3
2 2
3 3
样例输出
1 1
1 2
1 3
2 1
2 3
3 3
4 1
4 3

签到题。

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
#include<algorithm>
using namespace std;
#define INF 100861111
#define eps 1e-7
#define lson k*2
#define rson k*2+1
#define ll long long
int p1[55][55];
int p2[55][55];
int p3[55][55];
int main()
{
int i,j,k,n,m,x,y,test;
scanf("%d",&test);
while(test--)
{
scanf("%d%d",&n,&m);
memset(p1,0,sizeof(p1));
memset(p2,0,sizeof(p2));
memset(p3,0,sizeof(p3));
for(i=0;i<n;i++)
{
scanf("%d%d",&x,&y);
p1[x][y]=1;
}
for(i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
p2[x][y]=1;
}
for(i=0;i<=50;i++)
{
for(j=0;j<=50;j++)
{
for(k=0;k<=50;k++)
{
if(p1[i][k]&&p2[k][j])
p3[i][j]=1;
}
}
}
for(i=0;i<=50;i++)
{
for(j=0;j<=50;j++)
{
if(p3[i][j])
{
printf("%d %dn",i,j);
}
}
}
printf("n");
}
return 0;
}