# POJ 1050 To the Max C++版

OJ题解 5544浏览

DP，最大子段和，最大子矩阵和

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.

Sample Input

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

Sample Output

15

Slyar:给出n*n的矩阵，求和最大的子矩阵。

1、首先考虑一维的最大子段和问题，给出一个序列a[0]，a[1]，a[2]...a[n]，求出连续的一段，使其总和最大。

a[i]表示第i个元素
dp[i]表示以a[i]结尾的最大子段和

dp[i] = max{a[i], dp[i-1] + a[i]}

2、考虑二维的最大子矩阵问题

### 网友最新评论 (3)

1. 是的，合肥的那位朋友所指出的问题确实存在，将max初始化的时候设成数组的第一个元素即可
59snest11年前（2013-07-08）回复
2. 我发现你的程序貌似没考虑输出为负数的情况哈 输入4 -1 -2 -3 -4 输出 -1 但你的程序输出是0哦，改一下吧。^-^
匿名12年前（2012-07-13）回复
3. 讲解的很好，谢谢
匿名14年前（2010-04-11）回复