2017 ACM-ICPC 亚洲区-banana

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概述

Discussion

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 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.

Sample Input

1
6 4
1 1
1 2
2 1
2 3
3 3
4 1
1 1
1 3
2 2
3 3

Sample Output

1 1
1 2
1 3
2 1
2 2
2 3
3 3
4 1
4 3

题意：

输入n和m，n行每行表示第i只猴子喜欢第j种香蕉，m行每行表示第i种香蕉来自第j个地区；

问哪些猴子在哪些地区有喜欢的香蕉，按猴子序号（若同则按地区）升序输出。

思路：

通过观察，可把每对数据当作一个二元关系，由n行部分和m行部分的二元关系通过关系复合（离散数学）得到结果。这里可通过 vector 和 map 的实现。

code：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

const int maxn = 55;
vector<int >nm[maxn];
int map[maxn][maxn];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n, m, moky, bana, pls;
        for(int i=0;i<=50;i++)
            nm[i].clear();
        memset(map,0,sizeof(map));
        scanf("%d %d",&n, &m);
        for(int i = 0; i < n; i ++)
        {
            scanf("%d %d",&moky, &bana);
            nm[bana].push_back(moky);
        }
        for(int i = 0; i < m; i ++)
        {
            scanf("%d %d",&bana, &pls);
            for(unsigned int j = 0; j < nm[bana].size(); j ++)
                map[nm[bana][j]][pls] = 1;
        }
        for(int i = 1; i <= 50; i ++)
        {
            for(int j = 1; j <= 50; j ++)
                if(map[i][j])
                    printf("%d %dn",i,j);
        }
        printf("n");
    }
}