1139 First Contact (30分)--PAT甲级真题(测试点不通过来看看)

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:

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: 给出多对朋友的名单,如果ID前由负号"-",表示女孩子;
给出两个人a,b,要求找出所有可以介绍a和b认识的朋友对c,d;其中c和a为同性朋友,d和b为同性朋友;c和d是朋友;如果找不到,输出0

分析:在输入朋友对的时候,用vector存储每个人的同性朋友,便于之后查找;并且对于每一组输入a,b,用map来标记a,b是否为朋友关系;

易错点:
(1) c和d不能与a和b相同,即a和b本来就是好朋友,按照上面的方法寻找时,要排除这种情况;
(2)c和d不能是同一个人,即c既是a的朋友,又是b的朋友,要排除这种情况;
测试点2:输入的时候要按照字符串处理负号"-",不能当作数字用正负来判断,因为可能出现"-0000"这样子的数据;

#include 
#include 
#include
#include
#include 
using namespace std; 
vector<int> samSex[10010];//一个人同性朋友的集合
unordered_map<int, bool> judF;//两个人是否是朋友
int N, M, K; 
struct ans{
	int a, b;
	bool operator<(const ans& A)const {
		return (a != A.a) ? a < A.a : b < A.b;
	}
}; 
set<ans> ret;//查询对应的结果 
int main(){
	scanf("%d%d", &N, &M); 
	int a, b;
	for (int i = 0; i < M; i++){
		string strA, strB;
		cin >> strA >> strB; 
		int flagA = strA[0] == '-' ? -1 : 1;
		int flagB = strB[0] == '-' ? -1 : 1;
		a = strA[0] == '-' ? atoi(strA.substr(1, strA.length() - 1).c_str()) : atoi(strA.c_str());
		b = strB[0] == '-' ? atoi(strB.substr(1, strB.length() - 1).c_str()) : atoi(strB.c_str());
		if (flagA*flagB>0){//如果两个人是同性朋友，保存
			samSex[a].push_back(b);
			samSex[b].push_back(a);
		}
		judF[10000 * a + b] = true; //标志a和b是朋友关系
		judF[10000 * b + a] = true;
	} 
	scanf("%d", &K);  
	for (int i = 0; i < K; i++){
		ret.clear();
		scanf("%d%d", &a, &b);
		a = abs(a);
		b = abs(b);
		for (int i = 0; i < samSex[a].size(); i++){
			for (int j = 0; j < samSex[b].size(); j++){
				if (judF[10000 * samSex[a][i] + samSex[b][j]] == true &&
					samSex[a][i] != samSex[b][j] && samSex[a][i] != b && samSex[b][j] != a){
					ans tmp;
					tmp.a = samSex[a][i];
					tmp.b = samSex[b][j];
					ret.insert(tmp);
				}
			}
		}
		printf("%d\n", ret.size());
		for (auto it = ret.begin(); it != ret.end(); it++)
			printf("%04d %04d\n", it->a, it->b); 
	}  
	return 0;
}