1139 First Contact (30 分)

2019-05-11

1139 First Contact (30 分)

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:

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一堆人中的朋友关系。然后给出两人A，B。A要追B，必须找他的一个同性好友C，然后C要找B的一个同性好友D，并且CD是朋友关系。给出AB，问可能的中间人。

分析

因为只中转两次，所以没必要dfs之类的了。找到A，B的同性朋友，然后判断A的朋友和B的朋友是否认识即可。存储某人的朋友用邻接表，判断两人是否认识用邻接矩阵。

注意A可能直接认识B，所以要排除。然后输入会有-0000的，注意此时的性别判断。

代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>

using namespace std;
int n,m,k;
vector<int> r[10000];
struct node{
    int v1;
    int v2;
    node(int v1,int v2):v1(v1),v2(v2){}
};
vector<node> res;
int isfrds[10000][10000];
map<int,int> girl;
bool cmp(node n1,node n2){
    return n1.v1!=n2.v1?n1.v1<n2.v1:n1.v2<n2.v2;
}
int main() {
    scanf("%d %d",&n,&m);
    string s1,s2;
    int a,b;
    for(int i=0;i<m;i++){
        cin>>s1>>s2;
        if(s1[0]=='-'){
            a=stoi(s1.substr(1,s1.length()-1));
            girl[a]=1;
        }else
            a=stoi(s1);
        if(s2[0]=='-'){
            b=stoi(s2.substr(1,s2.length()-1));
            girl[b]=1;
        }else
            b=stoi(s2);
        r[a].push_back(b);
        r[b].push_back(a);
        isfrds[a][b]=isfrds[b][a]=1;
    }
    cin>>k;
    for(int i=0;i<k;i++){
        res.clear();
        scanf("%d %d",&a,&b);
        a=abs(a);
        b=abs(b);
        for(int afrds:r[a]){
            if(girl[a]==girl[afrds]&&afrds!=b){
                for(int bfrds:r[b]){
                    if(girl[b]==girl[bfrds]&&bfrds!=a){
                        if(isfrds[afrds][bfrds]==1){
                            node t(afrds,bfrds);
                            res.push_back(t);
                        }
                    }
                }
            }
        }
        sort(res.begin(),res.end(),cmp);
        cout<<res.size()<<endl;
        for(auto v:res){
            printf("%04d %04d\n",v.v1,v.v2);
        }
    }
    return 0;
}