Friend-Graph(中国大学生程序选拔赛2017年网络大赛)
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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.
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. Thei 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.
The first line od each case should contain one integers n, representing the number of people of the team.(
Then there are n-1 rows. The
Output
Please output ”Great Team!” if this team is a good team, otherwise please output “Bad Team!”.
Sample Input
141 1 00 01
Sample Output
Great Team!
分析:用数组存,然后找一个人与任意两个人伙伴或非伙伴,然后若这两个人为伙伴或非伙伴,则为不好的团体。
#include<bits/stdc++.h>using namespace std;int a[3005][3005];int main(){ int t; cin>>t; while(t--) { int n,z=0; cin>>n; for(int i=1;i<n;i++) for(int j=1;j<=n-i;j++) cin>>a[i][j]; for(int i=1;i<n;i++) { if(z==1) break; for(int j=1;j<=n-i;j++) { if(z==1) break; if(a[i][j]==1) { for(int l=j+1;l<=n-i;l++) if(a[i][l]==1&&a[j+i][l-j]==1) {z=1;break;} } if(a[i][j]==0) { for(int l=j+1;l<=n-i;l++) if(a[i][l]==0&&a[j+i][l-j]==0) {z=1;break;} } } } if(z==1) cout<<"Bad Team!"<<endl; else cout<<"Great Team!"<<endl; }}
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