hdu 1081 To The Max（最大子矩阵和）

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最新推荐文章于 2024-07-31 10:00:00 发布

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本文介绍了一种使用动态规划解决二维数组中最大子矩阵和问题的方法。通过将二维问题转化为一维问题，简化了求解过程。详细解释了如何在给定的数组中找到具有最大和的子矩阵，并提供了实现代码。

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To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7533 Accepted Submission(s): 3647

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

动态规划求最大子矩阵和，把二维的转化为一维的就好做了。

1、首先看一位数组a[]求最大字段和，b记录以i结尾的数组最大字段和；

显然若b<0（此时存储的是以i-1为结尾的最大和）,有b=a[i]; 否则b+=a[i];

2、转化为一维的数组就是枚举某一连续几行，把每一列的值加起来存到一个元素中，

这些元素就构成了一个一维数组。

然后，按照一维数组求最大字段和的方法求解即可。

#include"stdio.h"
#include"math.h"
#include"string.h"
#define N 105
int Maxsum(int a[],int m)
{                         //求一维数组的最大字段和
	int i,s=0,max=a[0];
	for(i=0;i<m;i++)
	{
		if(s>=0)
			s+=a[i];
		else
			s=a[i];
		if(s>max)
			max=s;
	}
	return max;
}
int main()
{
	int n,i,j,k;
	int b[N];
	int a[N][N];
	while(scanf("%d",&n)!=-1)
	{
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
				scanf("%d",&a[i][j]);
		int max=a[0][0];
		for(i=0;i<n;i++)    
		{
			memset(b,0,sizeof(b));
			for(j=i;j<n;j++)        //从某一行开始加到末行
			{
				for(k=0;k<n;k++)         //求解相加过程中的最大字段和
					b[k]+=a[j][k];
				int t=Maxsum(b,n);
				if(t>max)
					max=t;
			}
		}
		printf("%d\n",max);
	}
	return 0;
}