Maximum Sum

Background

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.

The Problem

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size  or greater located within the whole array. As an example, the maximal sub-rectangle of the array:

is in the lower-left-hand corner:

and has the sum of 15.

Input and Output

The input consists of an  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  integers separated by white-space (newlines and spaces). These  integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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].

The output is 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

C Code

#include<stdio.h>
#define Dif(i,j,k) (Table[i+k][j] - Table[i][j])
#define MAXN 110

int N, MAX;
int Table[MAXN][MAXN];


void ReadCase() {
   int i, j;
   for(i = 1; i<=N; i++)
	   for(j  = 0; j<N; j++)
		   scanf("%d",&Table[i][j]);
	   
}

void Cal() {
   int i, j, k, t;
   for(i = 1; i<=N; i++){
	   for(j = 0; j<N; j++)
		  Table[i][j] =  Table[i][j] + Table[i-1][j];
   }
   MAX = Table[1][0];
   for(k = 1; k<=N; k++) {
	   for(i = 0; i<=N-k; i++){
		   for(t = 0, j = 0; j<N; j++) {
		      if(t>=0) t+= Dif(i,j,k);
			  else t = Dif(i,j,k);
			  if(t>MAX) MAX = t;
		   }
	   }
   }
		 
   printf("%d\n",MAX);
   
}


void FREE() {
   int i, j;
   for(i = 0; i<=N; i++)
	   for(j = 0; j<=N; j++)
		   Table[i][j] = 0;
}

void main() {
    int f = 0;	

	while(scanf("%d",&N) == 1) {
	   if(f++) FREE();
	   ReadCase();
	   Cal();
	   
	}
}