1025：To the max（DP）

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;
}