2013
11-27

# To The Max

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.

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 the sum of the maximal sub-rectangle.

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

15

一道简单的求子矩阵的最大和问题，由于数据给的比较下，n只有100，可以使用常规的O(N^3)的时间复杂度的算法解，枚举每一行以及改行下面的所有的行的和，时间复杂度为O(N^2)，然后对压缩后的每一行进行dp，时间复杂度为线性的O(N)，最后总的时间复杂度为O(N^3)

dp的状态转移方程：dp[i] = MAX(dp[i-1] + tmp[i], dp[i])

#include <stdio.h>
#include <string.h>
#define MAX(x, y) ((x) > (y) ? (x) : (y))
int main(void)
{
int map[101][101];
int dp[101], tmp[101];
int max;
int i, j, t;
int n;

while(scanf("%d", &n) != EOF)
{
for(i = 0; i < n; i++)
for(j = 0; j < n; j++)
scanf("%d", &map[i][j]);
max = -10000000;
for(i = 0; i < n; i++)
{
memset(tmp, 0, sizeof(tmp));
for(t = i; t < n; t++)
{
for(j = 0; j < n; j++)
tmp[j] += map[t][j];

dp[0] = tmp[0];
for(j = 1; j < n; j++)
{
/*dp[j] = MAX(dp[j-1] + tmp[j], tmp[j]);
max = MAX(dp[j], max);*/
if(dp[j-1] > 0)
dp[j] = dp[j-1] + tmp[j];
else
dp[j] = tmp[j];
if(dp[j] > max)
max = dp[j];
}
}
}
printf("%d\n", max);
}
return 0;
}

1. Thanks for using the time to examine this, I truly feel strongly about it and enjoy finding out far more on this subject matter. If achievable, as you achieve knowledge