HDU 1081 To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10594    Accepted Submission(s): 5086

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

 

Source

Greater New York 2001

 

 

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn=200;
int a[maxn][maxn];

int main()
{
    int n,t,sum,max;
    while(scanf("%d",&n)!=EOF)
    {
        max=0;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                scanf("%d",&t);
                a[i][j]=a[i-1][j]+t;
            }
        }

        for(int i=1;i<n-1;i++){
            for(int j=i;j<=n;j++){
                sum=0;
                for(int k=1;k<=n;k++){
                    t=a[j][k]-a[i-1][k];
                    sum+=t;
                    if(sum<0) sum=0;
                    if(sum>max) max=sum; 
                }
            }
        }
        printf("%d
",max);
    }
}