Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array: 

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1 8 0 -2

is in the lower left corner: 

9 2

-4 1

-1 8

and has a sum of 15.

Input

The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

• Sample
  Input
• Raw

4

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1 8  0 -2

• Sample
  Output
• Raw

15

#include<cstdio>

#include<cstring>

using namespace std;

const int N=105;

int a[N][N];

int b[N];

int main ()

{

    int r,max=0;

    scanf("%d",&r);

    memset(a,0,sizeof(a));

    for(int i=1;i<=r;i++)

    {

        for(int j=1;j<=r;j++)

        {

            scanf("%d",&a[i][j]);

            a[i][j]+=a[i-1][j];

        }

    }

    max=a[1][1];

    for(int i=1;i<=r;i++)//从第i行开始到底j行。。。。转化成一维的

    {

        for(int j=i;j<=r;j++)

        {

            memset(b,0,sizeof(b));

            for(int k=1;k<=r;k++)

            {

                if(b[k-1]>=0)

                    b[k]=b[k-1]+a[j][k]-a[i-1][k];

                    else

                    b[k]=a[j][k]-a[i-1][k];

                if(max<b[k])

                    max=b[k];

            }

        }

    }

    printf("%d\n",max);

    return 0;

}