Sum Of SubRectangular Parallelepiped 三维最大子段和

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本文介绍了一种方法来计算矩形平行六面体中的最大子矩形平行六面体和，包括输入解析、求和矩阵构建及最大和计算。

Sum Of SubRectangular Parallelepiped

TimeLimit: 2 Second MemoryLimit: 32 Megabyte

Totalsubmit: 468 Accepted: 132

Description

There is a rectangular parallelepiped with a integer in each cell.Can you caculate the maximum sum of sub rectangular parallelepipeds?

Input

Input contains multiple test cases. The first line is the test number T, followed by cases. Each case begin with three integer L,M,N presenting the rectangular parallelepiped you will get. Then L matrixes each with M rows and N column follow.

Output

For each case print a line with the maximum sum of sub rectangular parallelepipeds.If all the integers are negtive,print 0.

Sample Input

1
2 2 2
7 -4
3 7

-9 -6
4 7

Sample Output

#include<iostream>
#include<cstring>
using namespace std;

int a[100][100][100];
int summ[100][100][100];
int i,j,k;
int l,m,n;
int t;

int main()
{
int i,j,k,p,q;
scanf("%d",&t);
while (t--)
{
  scanf("%d%d%d",&l,&m,&n);
  for (i=0;i<l;i++)
   for (j=0;j<m;j++)
    for (k=0;k<n;k++)
     scanf("%d",&a[i][j][k]);
  memset(summ,0,sizeof(summ));
  for (i=0;i<l;i++)
   for (j=0;j<m;j++)
       for (k=0;k<n;k++)
       {
     if (j==0&&k==0)
     {
      summ[i][j][k]=a[i][j][k];
      continue;
     }
     if (j==0)
     {
      summ[i][j][k]=summ[i][j][k-1]+a[i][j][k];
      continue;
     }
     if (k==0)
     {
      summ[i][j][k]=summ[i][j-1][k]+a[i][j][k];
      continue;
     }
     summ[i][j][k]=summ[i][j-1][k]+summ[i][j][k-1]-summ[i][j-1][k-1]+a[i][j][k];
    }
  int maxx=0;
  for (i=0;i<m;i++)
   for (j=0;j<n;j++)
    for (k=i;k<m;k++)
     for (p=j;p<n;p++)
     {
      int s=0;
      for (q=0;q<l;q++)
      {
       int a;
       if (i==0&&j==0)
        a=summ[q][k][p];
       else if (i==0)
        a=summ[q][k][p]-summ[q][k][j-1];
       else if (j==0)
        a=summ[q][k][p]-summ[q][i-1][p];
       else
        a=summ[q][k][p]-summ[q][i-1][p]-summ[q][k][j-1]+summ[q][i-1][j-1];
       s+=a;
       if (s<0) s=0;
       else if(s>maxx) maxx=s;
      }
     }
  printf("%d/n",maxx);
}
return 0;
}