hdu 6152 Friend-Graph

Friend-Graph

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 968    Accepted Submission(s): 507

Problem Description

It is well known that small groups are not conducive of the development of a team. Therefore, there shouldn’t be any small groups in a good team.
In a team with n members,if there are three or more members are not friends with each other or there are three or more members who are friends with each other. The team meeting the above conditions can be called a bad team.Otherwise,the team is a good team.
A company is going to make an assessment of each team in this company. We have known the team with n members and all the friend relationship among these n individuals. Please judge whether it is a good team.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.（T<=15）
The first line od each case should contain one integers n, representing the number of people of the team.( n≤3000)

Then there are n-1 rows. The ith row should contain n-i numbers, in which number aij represents the relationship between member i and member j+i. 0 means these two individuals are not friends. 1 means these two individuals are friends.

 

Output

Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.

 

Sample Input

1

4

1 1 0

0 0

1

 

Sample Output

Great Team!

 

Source

2017中国大学生程序设计竞赛 - 网络选拔赛

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题解大意：

判断这n个人组成的是一个great team还是bad team，

如果在这n个人中有3人及以上相互不认识货值有三人及以上相互认识，则是bad team，否则great team

题解：

推出，一个点最多连两个点，当点数>=6，要把这些点连起来，必然变成bad team

前五个暴力即可，这题当时mle了两次，下次要注意开的数组大小和题目中的限制

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

bool a[3002][3002];
int t,n;

bool work()
{
   for(int i=1;i<=n;i++)
   {
       for(int j=i+1;j<=n;j++)
        if(a[i][j]==1)
        {
            for(int k=j+1;k<=n;k++)
                if (a[j][k]==1 && a[k][i]==1) return 0;
        } else
        {
            for(int k=j+1;k<=n;k++)
                if(a[j][k]==0 && a[k][i]==0) return 0;
        }
   }
   return 1;
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n-i;j++)
            {
                int x;
                scanf("%d",&x);
                a[i][i+j]=x;
                a[i+j][i]=x;
            }
        }
        if (n>5) {printf("Bad Team!\n"); continue;}
      if(work()) printf("Great Team!\n");
        else printf("Bad Team!\n");
    }
  return 0;
}

 

转载于:https://www.cnblogs.com/stepping/p/7401080.html