PAT甲级 1139

最新推荐文章于 2021-02-07 18:47:04 发布

原创最新推荐文章于 2021-02-07 18:47:04 发布 · 305 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT

本文探讨了一个有趣的算法问题，即在一个友谊网络中，如何为一位想要追求心仪对象的人找到合适的中间人，通过一系列复杂的关系链，实现两人的初次接触。问题涉及到图论中的查找路径和匹配算法，同时考虑了性别匹配和直接联系的排除条件。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:

题意：A如果想和B在一起，那他就要先找自己的同性朋友C,再找B的同性朋友D，如果D和C是朋友，则输出C和D；如果A的同性朋友是B或者B的同性朋友是A是要避免输出这种结果。

这道题我并没有的到满分，系统提示超时~~~~~

#include <iostream>
#include<cstring>
#include<cmath>
#include<stdlib.h>
#include<string>
#include<vector>
#include<algorithm>
#include<bits/stdc++.h>
using namespace std;
bool gender[10001],G[10001][10001];
vector <vector<int>> G2(10001);
struct edge{
int f1,f2;};
bool cmp(edge a,edge b){
if(a.f1!=b.f1) return a.f1<b.f1;
 return a.f2<b.f2;

}
int main()
{
   int n,m,k;
   cin>>n>>m;
   int i,u,v;
   string s1,s2;
   for(i=0;i<m;i++){
    cin>>s1>>s2;
    u=abs(stoi(s1));
    v=abs(stoi(s2));
    if(s1[0]=='-') gender[u]=true;
    if(s2[0]=='-') gender[v]=true;
    G[u][v]=true;
    G[v][u]=true;
    G2[u].push_back(v);
    G2[v].push_back(u);

       }
       cin>>k;
       for(i=0;i<k;i++){
        cin>>u>>v;
        u=abs(u);
        v=abs(v);
        vector<edge> vectice;
        for(int x:G2[u]){
            if(x==v) continue;
            if(!G[u][x]||(gender[x]!=gender[u]))  continue;
            for(int y:G2[x]){
                if(y==u)continue;
                if(G[y][x]&&G[y][v]&&gender[y]==gender[v]){
                    edge z;
                    z.f1=x;
                    z.f2=y;
                    vectice.push_back(z);

                }
            }
        }

        sort(vectice.begin(),vectice.end(),cmp);
        cout<<vectice.size()<<endl;
        for(edge h:vectice)
            printf("%04d %04d\n",h.f1,h.f2);

               }
    return 0;
}