HDU - 6152 Friend-Graph(暴力)
2024-09-02 14:48:42
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;
}
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