HDU 6152 Friend-Graph （CCPC2017）

Friend-Graph

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1397    Accepted Submission(s): 710

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

141 1 00 01

 

Sample Output

Great Team!

题意：存在三个人包括三人以上互相友好或不友好就是坏队，其他的都是好队

转化成找三角关系，存在由三个1构成的三角或三个0构成的三角就是Bad，三个以上互相连接就一定包括三个互相连接，所以只用考虑三个就行

题目数据量不大，所以直接找也可以，枚举边，再通过边找点。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;bool a[3001][3001];int main(){    int t,n,i,j,k;    bool e;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=1; i<n; i++)        {            for(j=i+1; j<=n; j++)            {                scanf("%d",&a[i][j]);            }        }        e=0;        for(i=1; i<n; i++)        {            for(j=i+1; j<=n; j++)            {                if(a[i][j])                {                    for(k=j+1; k<=n; k++)                        if(a[i][k]&&a[j][k])                        {                            e=1;                            goto End;                        }                }                else                {                    for(k=j+1; k<=n; k++)                        if(!a[i][k]&&!a[j][k])                        {                            e=1;                            goto End;                        }                }            }        }End:        if(e)            printf("Bad Team!\n");        else            printf("Great Team!\n");    }    return 0;}