To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7927    Accepted Submission(s): 3831

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

4

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1
8 0 -2

Sample Output

15

二维最大连续子段和，枚举列数（i->j)第i列到第j列，计算每一行第i列到第j列的和，然后用每一行的这个和用一维的求最大连续子序列的和的方法求就行了，

求和的时候可以预处理一下，ma[i][j]表示第i行前j列的和，这样求第i列到第j列的和就可以用ma[k][j] - ma[k][i-1]表示。

#include <stdio.h>
#include <string.h>
#define N 105
#define INF 0x3f3f3f3f
int ma[N][N];
int main()
{
    int n,i,j,k,maxx,sum;
    while(~scanf("%d",&n))
    {
        memset(ma,0,sizeof(ma));
        for(i = 1; i<= n ; i++)
        {
            for(j = 1 ; j<= n ;j++)
            {
                scanf("%d",&ma[i][j]);
                ma[i][j]+=ma[i][j-1];
            }
        }
        sum = 0 ;maxx = -INF;
        for(i = 1 ; i <= n ; i++)
        {
            for(j = i ; j <= n ; j++)
            {
                //printf("%d ",ma[i][j]);
                sum = 0;
                for(k = 1 ; k<= n ; k++)
                {
                    //if(sum==17) printf("i = %d j = %d k = %d\n",i,j,k);
                    sum+=ma[k][j] - ma[k][i-1];
                    if(sum>maxx) maxx = sum;
                    if(sum<0) sum = 0;
                }
            }
           // printf("\n");
        }
        printf("%d\n",maxx);
    }
    return 0;
}

hdu 1081