HDU-4618 Palindrome Sub-Array 暴力枚举

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4618

　　直接暴力枚举中心点，在中间如果求不出最大值直接跳过优化下。。。

 //STATUS:C++_AC_31MS_800KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 #pragma comment(linker,"/STACK:102400000,102400000")
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=,M=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int w[N][N];
 int T,n,m,ans;

 int slove(int x,int y)
 {
     int i,j,c,ok,ret=;
     for(c=ok=;x-c>=&&x+c<n && y-c>=&&y+c<m;c++){
         for(i=;i<=c;i++){
             if(w[x-c+i][y-c]!=w[x-c+i][y+c]
                || w[x-c][y-c+i]!=w[x+c][y-c+i]
                || w[x+i][y-c]!=w[x-i][y-c]
                || w[x+i][y+c]!=w[x-i][y+c]
                || w[x-c][y-i]!=w[x-c][y+i]
                || w[x+c][y-i]!=w[x+c][y+i]){ok=;break;}
         }
         if(!ok)break;
     }
     ret=(c-)<<|;
     for(c=ok=;x-c+>=&&x+c<n && y-c+>=&&y+c<m;c++){
         for(i=;i<=c;i++){
             if(w[x-c+i+][y-c+]!=w[x-c+i+][y+c]
                || w[x-c+][y-c+i+]!=w[x+c][y-c+i+]
                || w[x+i][y-c+]!=w[x-i+][y-c+]
                || w[x+i][y+c]!=w[x-i+][y+c]
                || w[x-c+][y-i+]!=w[x-c+][y+i]
                || w[x+c][y-i+]!=w[x+c][y+i]){ok=;break;}
         }
         if(!ok)break;
     }
     return Max(ret,(c-)<<);
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         for(i=;i<n;i++){
             for(j=;j<m;j++){
                 scanf("%d",&w[i][j]);
             }
         }

         ans=;
         for(i=;i<n;i++){
             if( ((i+)<<)<=ans || (((n-i-)<<)|)<=ans)continue;
             for(j=;j<m;j++){
                 if( ((j+)<<)<=ans || (((m-j-)<<)|)<=ans)continue;
                     ans=Max(ans,slove(i,j));
             }
         }

         printf("%d\n",ans);
     }
     return ;
 }