POJ1050

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32838		Accepted: 17175

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

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

Sample Output

15

Source

Greater New York 2001

 

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int main()
{
    int n;
    int array[110][110];
    int dp[110][110];
    int sum, term, maxSum;

    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&array[i][j]);
            }
        }
        //dp
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                dp[i][j]=dp[i][j-1]+array[i][j];
            }
        }
        maxSum=-65535;
        for(int i=1;i<=n;i++)
        {
            for(int j=i; j<=n;j++)
            {
                sum=-1;
                for(int k=1;k<=n;k++)
                {
                    term=dp[k][j]-dp[k][i-1];
                    if(sum<0)  //sum[i]=max{sum[i-1]+a[i],a[i]}
                    {
                        sum=term;
                    }
                    else
                    {
                        sum+=term;
                    }
                    if(sum>maxSum)
                    {
                        maxSum=sum;
                    }
                }
            }
        }
        printf("%d\n",maxSum);
    }

    return 0;
}