G - To the Max POJ - 1050

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

最大子矩阵和问题，可以转化为连续行的最大子段和，也就是相当于一个最大字段和的升级版，实际上动态规划也必定是考虑到了所有的可能情况，但只是把一些子问题的解答保存下来，其实我认为这个最大子矩阵和问题，虽然是用动态规划但实际上并没有用到先前的子问题的解答，可以想象，一个子矩阵必定是连续行连续列所构成的，我们大可以将其简化为，不相同的连续行取不相同的连续列得到的最大和。

可能这段话有些绕口，可以看看以下这个矩阵。

a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34

a41 a42 a43 a44

如果我们要取一个2*n的矩阵，显然，连续的两行有n - 1种选择，也可以看到，这个列也是连续的，例如，我们取n = 1，且从第二第三行中选取，那么取一列的话有n种选择，且要从中找到最大的和，然后n = 2时，则又有n - 1种选择，我们可以注意到一个问题，当我们增加列数的时候，选取的每一行的那一列都必须加进去，进而我们可以想到，既然增加列数，每一行的那一列的数都必须加上去，那么我们大可以直接压缩一下状态，把选取的i行压缩成为1行，然后在这一行里面求一个最大字段和。

代码如下：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <string>
#include <cstring>
#define INF 0x3f3f3f3f
int a[105][105];
int main()
{
int N;
while (scanf("%d", &N) != EOF) {
int x;
for (int i = 1; i <= N; ++i) {
for (int j = 1; j <= N; ++j) {
scanf("%d", &x);
a[i][j] = a[i - 1][j] + x;
}
}
int MAX = -INF;
for (int i = 0; i <= N; ++i) {//对每一个可能的压缩的状态求最大字段和
for (int j = i + 1; j <= N; ++j) {
int sum = 0;
for (int k = 1; k <= N; ++k) {
sum = sum + a[j][k] - a[i][k];
MAX = std::max(sum, MAX);
if (sum < 0)
sum = 0;
}
}
}
printf("%d\n", MAX);
}
return 0;
}