2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛 A. Banana（连通性水题）

忘是亡心i 2022-06-08 00:45 211阅读 0赞

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 TT, indicating that there are TT test cases.

For each test case, the first line contains two integers NN and MM, representing the length of the first and the second lists respectively.

In the each line of following NN lines, two positive integers i, ji,j indicate that the ii-th monkey favours the jj-th type of banana.

In the each line of following MM 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 5050.

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

题解：

上次网赛只a了一题。。这次抱着只a 2，3题的心态去做，这次居然a出了五题hhhhh，不过还是膜拜学长们。。。帮我买学校搞到排名17orz

思路：

类似于弗洛伊德的思想，比较水，数据也很小

代码：

#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 %d\n",i,j);
                }
            }
        }
        printf("\n");
    
        }
        return 0;
    }