HDU6152

Posted Penn000

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了HDU6152相关的知识,希望对你有一定的参考价值。

Friend-Graph

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28    Accepted Submission(s): 15


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.(n3000)

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

 
三点不能成环,暴力判环
 1 //2017-08-19
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 const int N = 3050;
10 bool G[N][N];
11 int n;
12 
13 void work(){
14     for(int i = 1; i <= n; i++){
15         for(int j = i+1; j <= n; j++){
16             if(G[i][j] == 1){
17                 for(int k = j+1; k <= n; k++){
18                     if(G[i][k] == 1 && G[j][k] == 1){
19                         printf("Bad Team!\n");return;
20                     }
21                 }
22             }
23             if(G[i][j] == 0){
24                 for(int k = j+1; k <= n; k++){
25                     if(G[i][k] == 0 && G[j][k] == 0){
26                         printf("Bad Team!\n");return;
27                     }
28                 }
29             }
30         }
31     }
32     printf("Great Team!\n");
33 }
34 
35 int main()
36 {
37     int T, a;
38     scanf("%d", &T);
39     while(T--){
40         scanf("%d", &n);
41         for(int i = 1; i <= n-1; i++){
42             for(int j = i+1; j <= n; j++){
43                 scanf("%d", &a);
44                 G[j][i] = G[i][j] = a;
45             }
46         }
47         work();
48     }
49 
50     return 0;
51 }

 

以上是关于HDU6152的主要内容,如果未能解决你的问题,请参考以下文章

hdu6152 拉姆齐定理

HDU 6152 Friend-Graph

hdu 6152 Friend-Graph

HDU - 6152 Friend-Graph(暴力)

hdu 6152 : Friend-Graph (2017 CCPC网络赛 1003)

[HDOJ6152] Friend-Graph(拉姆齐定理)