HDU-6152 Friend-Graph （思维+暴力）

本文链接：https://blog.youkuaiyun.com/HHH_go_/article/details/77461184

本文介绍了一个算法，用于判断一个由n个人组成的团队是否为好团队。通过输入成员间的朋友关系，算法检查是否存在三名成员相互间全部是朋友或全部不是朋友的情况。若存在，则认为该团队不好。

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Friend-Graph

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1274 Accepted Submission(s): 657

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

i th 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

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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题意：

有n个人给出m个关系，如果任选3个人他们之间都不是朋友或者他们三个互相都是朋友，那么输出bad team 否则为 great team。

分析：

画图可以很容易知道，当n大于6时必定为 bad team （满足第一个条件那么不相邻的点要有连边，这样会导致连续的3个点会形成三角形）

那么当n小于等于6的时候直接暴力就可以啦~~

AC代码：

#include<stdio.h>
#include<string.h>
bool map[10][10];
bool solve(int n)
{
    for(int i=1;i<n;i++)
    {
	for(int j=i+1;j<n;j++)
	{
	    for(int k=j+1;k<=n;k++)
	    {
		if(map[i][j]==map[i][k]&&map[i][j]==map[j][k]&&map[j][k]==map[i][k])
		    return false;
	    }
	}
    }
    return true;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
	int n;
	scanf("%d",&n);
	if(n>6)
	{
	    for(int i=1;i<n;i++)
	    {
		for(int j=i+1;j<n;j++)
		{
		    int m;
		    scanf("%d",&m);
		}
	    }
	    printf("Bad Team!\n");
	}
	else
	{
	    memset(map,false,sizeof(false));
	    for(int i=1;i<n;i++)
	    {
		for(int j=i+1;j<=n;j++)
		{
		    int m;
		    scanf("%d",&m);
		    if(m)
			map[i][j]=map[j][i]=true;
		    else
			map[i][j]=map[j][i]=false;
		}
	    }
	    if(solve(n))
		printf("Great Team!\n");
	    else
		printf("Bad Team!\n");
	}
    }
}