Friend-Graph
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6514 Accepted Submission(s): 1610
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.
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.
#include<cstdio>
#include<cstring>
int a[1500][3000],b[3010],c[3010];
int main(){
int t,n,i,j,k,p,r;
scanf("%d",&t);
while(t--){
scanf("%d",&n);memset(a,0,sizeof(a));
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
if(i<1500)scanf("%d",&a[i][j]);
else scanf("%d",&a[3000-i][3000-j]);
}
}
int flag=0,x,y;
if(n<=2){
printf("Great Team!\n");continue;
}
if(n==3)if(a[1][2]&&a[1][3]&&a[2][3]){
printf("Great Team!\n");continue;
}
for(i=1;i<n;i++){
k=0;p=0;
if(i<1500)
for(j=i+1;j<=n;j++){
if(a[i][j])b[k++]=j;else c[p++]=j;
}
else
for(j=i+1;j<=n;j++){
if(a[3000-i][3000-j])b[k++]=j;else c[p++]=j;
}
for(j=0;j<k;j++){
for(r=j+1;r<k;r++){
if(b[j]<1500)x=b[j],y=b[r];
else x=3000-b[j],y=3000-b[r];
if(a[x][y]){flag++;break;}
}
if(flag)break;
}
if(flag)break;
for(j=0;j<p;j++){
for(r=j+1;r<p;r++){
if(b[j]<1500)x=c[j],y=c[r];
else x=3000-c[j],y=3000-c[r];
if(a[x][y]==0){flag++;break;}
}
if(flag)break;
}
if(flag)break;
}
if(flag)printf("Bad Team!\n");
else printf("Great Team!\n");
}
}