hdu 1081 To The Max（最大子矩阵和）

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

15

 

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=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;imax)
					max=t;
			}
		}
		printf("%d\n",max);
	}
	return 0;
}