PAT 1139 First Contact

原题链接：1139 First Contact (30分)
关键字：图论、枚举
参考的柳神博客：PAT 1139. First Contact (30)-PAT甲级真题

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（非递减） 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:

4
1002 2004
1003 2002
1003 2003
1004 2002
4
2001 1002
2001 1003
2002 1003
2002 1004
0
1
2003 2001
0

题目大意： 给出很多对朋友关系，问你是否能找到满足帮a和b“连线”的朋友。
也就是说是否存在满足A—— C—— D—— B（AC同性朋友，CD异性朋友，DB同性朋友）这样的可能。不满足的话输出0，满足的话输出所有可能的C、D(此处不需要负号)

分析:

• .用数组unordered_map arr 标记两个人是否是朋友（邻接矩阵表示）；
• 用 vector v[10000]标记所有人的同性朋友（邻接表表示），便于之后查找；
• 对于一对想要在一起的A和B，他们需要先找A的所有同性朋友C，再找B的所有同性朋友D，当C和D两人是朋友的时候则可以输出C和D
• A和B直接就是朋友的情况需要舍弃
• 如果用int接收一对朋友，-0000和0000对于int来说都是0，将无法得知这个人的性别，也就会影响他找他的同性朋友的那个步骤，所以考虑用字符串接收，因为题目中已经表示会以符号位加四位的方式给出输入，所以只要判断字符串是否长度相等，如果大小相等说明这两个人是同性
• 需要对结果数组ans进行排序

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
unordered_map<int, bool> arr;   //记录两个人是否是朋友（邻接矩阵）
struct node {
    int a, b;
};

bool cmp(node x, node y) {  //按a递增排序，如果a相同按b递增排序
    return x.a != y.a ? x.a < y.a : x.b < y.b;
}

int main() {
    int n, m, k;    //n总人数 m友好关系数 k询问数
    scanf("%d%d", &n, &m);
    vector<int> v[10000];   //标记所有人的同性朋友（邻接表）
    for (int i = 0; i < m; i++) {
        string a, b;
        cin >> a >> b;
        if (a.length() == b.length()) { //同性朋友关系存到v
            v[abs(stoi(a))].push_back(abs(stoi(b)));
            v[abs(stoi(b))].push_back(abs(stoi(a)));
        }
        arr[abs(stoi(a)) * 10000 + abs(stoi(b))] = arr[abs(stoi(b)) * 10000 + abs(stoi(a))] = true;	//异性朋友关系
    }
    scanf("%d", &k);
    for (int i = 0; i < k; i++) {   //k个询问
        int c, d;
        cin >> c >> d;	//其实就是题目问的A、B
        vector<node> ans;   //结果数组
        for (int j = 0; j < v[abs(c)].size(); j++) {
            for (int k = 0; k < v[abs(d)].size(); k++) {
                if (v[abs(c)][j] == abs(d) || abs(c) == v[abs(d)][k]) continue;	//A、B是同性朋友，跳过
                if (arr[v[abs(c)][j] * 10000 + v[abs(d)][k]] == true)	//A的同性朋友C和B的同性朋友D是朋友，存入结果集
                    ans.push_back(node{v[abs(c)][j], v[abs(d)][k]});
            }
        }
        sort(ans.begin(), ans.end(), cmp);
        printf("%d\n", int(ans.size()));
        for(int j = 0; j < ans.size(); j++)
            printf("%04d %04d\n", ans[j].a, ans[j].b);	//输出时以四位数的方式输出
    }
    return 0;
}