HDU - 6152 Friend-Graph（暴力）

题意：给定n个人的关系，若存在三个及以上的人两两友好或两两不友好，则"Bad Team!"，否则"Great Team!"。

分析：3000*3000内存10000+，因此存关系要用bool数组，否则mle。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-12;
inline int dcmp(double a, double b)
{
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 3000 + 10;
const int MAXT = 3025 + 10;
using namespace std;
bool pic[MAXN][MAXN];
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        memset(pic, 0, sizeof pic);
        int n;
        scanf("%d", &n);
        int x;
        for(int i = 1; i <= n; ++i){
            for(int j = i + 1; j <= n; ++j){
                scanf("%d", &x);
                if(x) pic[j][i] = pic[i][j] = true;
                else pic[j][i] = pic[i][j] = false;
            }
        }
        bool ok = true;
        for(int i = 1; i <= n; ++i){
            for(int j = i + 1; j <= n; ++j){
                for(int k = j + 1; k <= n; ++k){
                    if((pic[i][j] && pic[i][k] && pic[j][k]) || (!pic[i][j] && !pic[i][k] && !pic[j][k])){
                        ok = false;
                        break;
                    }
                }
                if(!ok) break;
            }
            if(!ok) break;
        }
        if(ok) printf("Great Team!\n");
        else printf("Bad Team!\n");
    }
    return 0;
}