HDOJ题目1081 To The Max（动态规划）

最新推荐文章于 2020-02-12 14:43:20 发布

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最新推荐文章于 2020-02-12 14:43:20 发布

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本文介绍了一个关于寻找二维数组中最大子矩阵和的问题，并提供了一段AC代码实现。该问题要求从给定的整数矩阵中找到具有最大和的连续子矩阵。

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To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8370 Accepted Submission(s): 4064

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

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

Sample Output

Source

Greater New York 2001

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ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
//#define INF 0xfffffff
int map[1010][1010];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int i,j,k,ans=-1e9;
		memset(map,0,sizeof(map));
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				int x;
				scanf("%d",&x);
				map[i][j]=map[i][j-1]+x;
			}
		}
		for(i=1;i<=n;i++)
		{
			//int temp=-1;
			for(j=1;j<=i;j++)
			{
				int temp=-1;
				for(k=1;k<=n;k++)
				{
					temp=max(temp,0)+map[k][i]-map[k][j-1];
					ans=max(ans,temp);
				}
			}
		}
		printf("%d\n",ans);
	}
}