HDU - 6152 Friend-Graph(暴力)
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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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hdu 6152 : Friend-Graph (2017 CCPC网络赛 1003)