Friend-Graph（HDU 3152）

Friend-Graph

[align=center]Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1752    Accepted Submission(s): 870
[/align]

[align=left]Problem Description[/align]

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.

 

[align=left]Input[/align]

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
4000
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.

 

[align=left]Output[/align]

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

 

[align=left]Sample Input[/align]

1
4
1 1 0
0 0
1

 

[align=left]Sample Output[/align]

Great Team!

 
//题目：有3个人以上互相认识或互相不认识就是Bad Team，不然就是Good Team。给定一张图，求Good
Team/Bad Team。

//思路：一开始用最大团做，不出意外的TLE了。后来发现暴力搜整张图中是否存在节点个数为3的环即可，图正反各求一遍。PS：只能用一个二维数组，否则会MLE。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

const short MAX=3005;

int n;
bool map[MAX][MAX];
bool vis[MAX];

//搜是否存在大小为3的环
bool judge()
{
memset(vis,false,sizeof(vis));
int i,j;
queue<int>q;
q.push(1);
vis[1]=true;
while(!q.empty())
{
int now=q.front();
q.pop();
for(i=1;i<n;i++)
{
for(j=i+1;j<=n;j++)
{
if(map[i][j]==true&&map[now][i]==true&&map[now][j]==true)
{
return false;
}
}
if(map[i][j]==true&&vis[i]==false)
{
vis[i]=true;
q.push(i);
}
}
}
return true;
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1;i<n;i++)
{
for(int j=i+1;j<=n;j++)
{
int num;
scanf("%d",&num);
if(num==1)
map[i][j]=map[j][i]=true;
else
map[i][j]=map[j][i]=false;
}
}
if(judge()==false)
{
printf("Bad Team!\n");
continue;
}
//建反图
for(int i=1;i<n;i++)
{
for(int j=i+1;j<=n;j++)
{
if(map[i][j]==true)
{
map[i][j]=map[j][i]=false;
}
else
{
map[i][j]=map[j][i]=true;
}
}
}
if(judge()==false)
printf("Bad Team!\n");
else
printf("Great Team!\n");
}
return 0;
}