PAT甲级——A1139 First Contact【30】

Unlike in nowadays, the way that boys and girls expressing their feelings of love was quite subtle in the early years. When a boy A had a crush on a girl B, he would usually not contact her directly in the first place. Instead, he might ask another boy C, one of his close friends, to ask another girl D, who was a friend of both B and C, to send a message to B -- quite a long shot, isn't it? Girls would do analogously.

Here given a network of friendship relations, you are supposed to help a boy or a girl to list all their friends who can possibly help them making the first contact.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (1 < N ≤ 300) and M, being the total number of people and the number of friendship relations, respectively. Then M lines follow, each gives a pair of friends. Here a person is represented by a 4-digit ID. To tell their genders, we use a negative sign to represent girls.

After the relations, a positive integer K (≤ 100) is given, which is the number of queries. Then K lines of queries follow, each gives a pair of lovers, separated by a space. It is assumed that the first one is having a crush on the second one.

Output Specification:

For each query, first print in a line the number of different pairs of friends they can find to help them, then in each line print the IDs of a pair of friends.

If the lovers A and B are of opposite genders, you must first print the friend of A who is of the same gender of A, then the friend of B, who is of the same gender of B. If they are of the same gender, then both friends must be in the same gender as theirs. It is guaranteed that each person has only one gender.

The friends must be printed in non-decreasing order of the first IDs, and for the same first ones, in increasing order of the seconds ones.

Sample Input:

10 18

-2001 1001

-2002 -2001

1004 1001

-2004 -2001

-2003 1005

1005 -2001

1001 -2003

1002 1001

1002 -2004

-2004 1001

1003 -2002

-2003 1003

1004 -2002

-2001 -2003

1001 1003

1003 -2001

1002 -2001

-2002 -2003

5

1001 -2001

-2003 1001

1005 -2001

-2002 -2004

1111 -2003

Sample Output:

 #include <iostream>

 #include <vector>

 #include <string>

 #include <algorithm>

 using namespace std;

 vector<int>graph[];//图

 bool boy[];//下标为ID，元素表示该ID是否是男

 int N, M, K, vstart, vend;

 int main()

 {

     scanf("%d%d", &N, &M);

     for (int i = ; i < M; ++i)

     {//读取边的数据

         string a, b;

         cin >> a >> b;

         int ia = abs(stoi(a)), ib = abs(stoi(b));//将字符串化为正整数

         graph[ia].push_back(ib);//向图中加入边

         graph[ib].push_back(ia);//向图中加入边

         boy[ia] = (a[] != '-');//表示该人性别

         boy[ib] = (b[] != '-');//表示该人性别

     }

     scanf("%d", &K);

     while (K--)

     {

         scanf("%d%d", &vstart, &vend);//读取首尾结点

         vector<pair<int, int>>result;//存储符合题目要求的两个结点

         for (int i : graph[abs(vstart)])//遍历首节点的朋友

             if (i != abs(vend) && i != abs(vstart) && boy[i] == boy[abs(vstart)])//找到非首尾结点且与首节点性别相同的朋友作为第一个节点

                 for (int j : graph[i])//遍历第一个节点的朋友

                     if (j != abs(vend) && j != abs(vstart) && boy[j] == boy[abs(vend)])//找到非首尾结点且与尾节点性别相同的朋友作为第二个节点

                         for (int k : graph[j])//遍历第二个节点的朋友

                             if (k == abs(vend))//尾结点是第二个节点的朋友

                                 result.push_back({ i,j });//i,j两个节点符合要求

         printf("%d\n", result.size());

         sort(result.begin(), result.end());//排序

         for (auto&i : result)//输出

             printf("%04d %04d\n", i.first, i.second);

     }

     return ;

 }