poj1050-To the Max

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.

主要思想就是降维，源代码如下：

#include <stdio.h>
#include <stdlib.h>
#define M 101
int num[M][M];
int N;

int submax(int a[M])
{
int i,pre=a[1],max=0;
for(i=2;i<=N;i++)
{
if(a[i]+pre>a[i])
pre=a[i]+pre;
else
pre=a[i];
if(pre>max)
max=pre;
}
return max;
}

int submax2()
{
int b[M];
int i,j,k,max=0;
for(i=1;i<=N;i++)
{
memset(b,0,sizeof(b));
for(j=i;j<=N;j++)
{
for(k=1;k<=N;k++)
b[k]+=num[j][k];
int ff=submax(b);
if(ff>max) max=ff;
}
}
return max;
}

int main()
{
int i,j,k;
scanf("%d",&N);
for(i=1;i<=N;i++)
for(j=1;j<=N;j++)
scanf("%d",&num[i][j]);
printf("%d
",submax2());

return 0;
}