hdu 6152 Friend"/>
hdu 6152 Friend
Friend-Graph
题目链接:Friend-Graph
题意:给出n个人的关系,如果其中有三个人(或三个以上)相互认识或相互不认识,就输出Bad Team!
,否则输出Great Team!
思路:拉姆齐定理
对于n≥6的情况,直接输出Bad Team!
n<6,直接暴力判断即可
代码:
#include<bits/stdc++.h>
using namespace std;const int maxn=3e3+5;
bool mp[maxn][maxn];bool judge(int n)
{for(int i=1; i<=n-2; ++i)for(int j=i+1; j<=n-1; ++j)for(int l=j+1; l<=n; ++l)if((mp[i][j]&&mp[i][l]&&mp[j][l])||(!mp[i][j]&&!mp[i][l]&&!mp[j][l]))return false;return true;
}int main()
{int t,n,x;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1; i<n; ++i)for(int j=i+1; j<=n; ++j){scanf("%d",&x);mp[i][j]=mp[j][i]=x;}if(n>=6){printf("Bad Team!\n");continue;}if(judge(n))printf("Great Team!\n");elseprintf("Bad Team!\n");}return 0;
}
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