UVa 10827 Maximum sum on a torus

UVa 10827 Maximum sum on a torus

最大子矩阵问题的增强版～难度增加了不少！但是我莫名其妙地一次AC了……AC之后发现某个地方和我原本设计的算法不符……修改到和我设计的算法符合……然后WA了（这难道就是传说中的RP么）。
我说说我AC的算法：首先，把原矩阵复制成4份，这样就解决了矩阵循环的问题了；然后按照基础最大子矩阵的方法枚举上下边，但是需要加入一个限制：上下边的距离不能超过n（这个很好理解）；上下边固定住以后，同样和基础最大子矩阵的做法一样：DP。同时记录当前序列的长度，长度如果等于n，从这个点到它前n个点的DP状态需要重新计算（相当于 重新选定了当前行的起点，这样会使算法复杂度增高到O(n^4)，不过长度等于n的情况太少啦，所以依然不会超时）。
后记：本人之前的算法还是可行的，只不过我代码不小心写错啦～具体和上面的不同之处就在于检测到长度等于n之后，计算当前点的状态不再使用递推，而改用枚举序列起点的方式，两种方法复杂度相同。
之前认为我本来的算法有误是这样想的：比如当前计算出一个[i,i+n-1]的序列，现在要计算i+n这一个点的状态。直接枚举[p,i+n]（其中i+1<=p<=i+n）这些序列即可。之前认为最优解可能出现在[i+1,p-1]这一段，所以需要重新从i+1开始递推，后来想想是不可能出现这种情况的。
枚举的方法修改成了如下代码：

if (len == n)
{
    now =- kInf;
    len = 0 ;
     for ( int  p = k - n + 1 ;p <= k;p ++ )
    {
         if (now < d[j][k] - d[j][p - 1 ] - d[i - 1 ][k] + d[i - 1 ][p - 1 ])
        {
            now = d[j][k] - d[j][p - 1 ] - d[i - 1 ][k] + d[i - 1 ][p - 1 ];
            len = k - p + 1 ;
        }
    }
    tmp = max(tmp,now);
}

以下是我的代码：

#include < algorithm >
#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxn( 77 );
const   int  kInf( 0x7f7f7f7f );

int  main()
{
    #ifndef ONLINE_JUDGE
    freopen( " data.in " , " r " ,stdin);
    freopen( " data.out " , " w " ,stdout);
     #endif

     int  T;
    scanf( " %d " , & T);
     while (T -- )
    {
         int  n;
        scanf( " %d " , & n);
         int  r[kMaxn << 1 ][kMaxn << 1 ];
         for ( int  i = 1 ;i <= n;i ++ )
             for ( int  j = 1 ;j <= n;j ++ )
                scanf( " %d " , & r[i][j]);
         for ( int  i = 1 ;i <= n;i ++ )
             for ( int  j = n + 1 ;j <= (n << 1 );j ++ )
                r[i][j] = r[i][j - n];
         for ( int  i = n + 1 ;i <= (n << 1 );i ++ )
             for ( int  j = 1 ;j <= n;j ++ )
                r[i][j] = r[i - n][j];
         for ( int  i = n + 1 ;i <= (n << 1 );i ++ )
             for ( int  j = n + 1 ;j <= (n << 1 );j ++ )
                r[i][j] = r[i - n][j - n];

         int  d[kMaxn << 1 ][kMaxn << 1 ];
        memset(d, 0 ,(kMaxn << 1 ) * (kMaxn << 1 ) * sizeof ( int ));
         for ( int  i = 1 ;i <= (n << 1 );i ++ )
             for ( int  j = 1 ;j <= (n << 1 );j ++ )
                d[i][j] = d[i - 1 ][j] + d[i][j - 1 ] - d[i - 1 ][j - 1 ] + r[i][j];

         int  ans( - kInf);
         for ( int  i = 1 ;i <= (n << 1 );i ++ )
             for ( int  j = i;j <= (n << 1 )  &&  j <= i + n - 1 ;j ++ )
            {
                 int  t[kMaxn << 1 ];
                 for ( int  k = 1 ;k <= (n << 1 );k ++ )
                    t[k] = d[j][k] - d[j][k - 1 ] - d[i - 1 ][k] + d[i - 1 ][k - 1 ];

                 int  tmp( - kInf),now( - kInf),len( 0 );
                 for ( int  k = 1 ;k <= (n << 1 );k ++ )
                {
                     if (len == n)
                    {
                        now =- kInf;
                        len = 0 ;
                         for ( int  p = k - n + 1 ;p <= k;p ++ )
                        {
                             if (now + t[p] > t[p])
                            {
                                now = now + t[p];
                                len ++ ;
                            }
                             else
                            {
                                now = t[p];
                                len = 1 ;
                            }
                            tmp = max(tmp,now);
                        }
                    }
                     else
                    {
                         if (now + t[k] > t[k])
                        {
                            now = now + t[k];
                            len ++ ;
                        }
                         else
                        {
                            now = t[k];
                            len = 1 ;
                        }
                        tmp = max(tmp,now);
                    }
                }
                ans = max(ans,tmp);
            }

        printf( " %d\n " ,ans);
    }

     return   0 ;
}