Friend-Graph
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3254 Accepted Submission(s): 523
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!
这道题当时题意是猜的自己真的没看懂,英语太烦了.
1 #include <bits/stdc++.h> 2 using namespace std; 3 int num[3010]; 4 int main(){ 5 int T; 6 scanf("%d",&T); 7 while(T--){ 8 int n; 9 scanf("%d",&n); 10 bool prime=true; 11 if(n>5) 12 prime=false; 13 int m; 14 memset(num,0,sizeof(num)); 15 for(int i=1;i<n;i++){ 16 for(int j=1;j<=n-i;j++){ 17 scanf("%d",&m); 18 if(m==1){ 19 num[i]++; 20 num[i+j]++; 21 } 22 } 23 } 24 if(!prime) 25 printf("Bad Team!\n"); 26 else{ 27 for(int i=1;i<=n;i++){ 28 if(num[i]>=3||(n-1-num[i])>=3) 29 prime=false; 30 } 31 if(prime) 32 printf("Great Team!\n"); 33 else 34 printf("Bad Team!\n"); 35 } 36 } 37 return 0; 38 }