hdu1081 To The Max--DP（最大子矩阵和）-白红宇的个人博客

发布日期：2021-10-03 20:32:02 浏览次数：2 分类：技术文章

本文共 1842 字，大约阅读时间需要 6 分钟。

原题链接：

一：原题内容

Problem Description

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.

Input

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

Output the sum of the maximal sub-rectangle.

Sample Input

40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2

Sample Output

二：分析理解

这道题前身是求一维数组的最大子序列和。这题只是二维而已。

三：AC代码

#include
   
      #include
    
     #include
     
        using namespace std;int map[110][110];int main(){	int n;	int x;	while (~scanf("%d", &n))	{		memset(map, 0, sizeof(map));		for (int i = 1; i <= n; i++)			for (int j = 1; j <= n; j++)			{				scanf("%d", &x);				map[i][j] += (x + map[i][j - 1]);			}		int ans = -1000000;		for (int i = 1; i <= n; i++)//第i列		{			for (int j = i; j <= n; j++)//第j列			{				int maxx = 0;				for (int k = 1; k <= n; k++)//第k行				{					if (maxx < 0)						maxx = 0;					maxx += (map[k][j] - map[k][i - 1]);					if (maxx > ans)						ans = maxx;				}			}		}		printf("%d\n", ans);	}	return 0;}

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