二维数组的子数组之和的最大值

对于一维的数组，要求其子数组之和的最大值很简单，动态规划轻松解决，复杂度O(N)  

代码如下：

#include <iostream>
using namespace std;
int MaxSum(int a[],int n)
{
	int start = a[0];
	int max = a[0];
	for(int i=1;i<n;i++)
	{
		if(start<0)
			start = 0;
		start += a[i];
		if(start>max)
			max = start;
	}
	return max;
}
int main()
{
	int a[] = {1,2,-4,3,5,7,-5,6};
	cout<<MaxSum(a,8);
	return 0;
}

代码可参看：

#include <iostream>
using namespace std;
#define M 5
#define N 5
int PS[M+1][N+1];
int BC(int a,int c,int i)
{
	return PS[c][i]-PS[a-1][i]-PS[c][i-1]+PS[a-1][i-1];
}
int MaxSum(int A[M][N])
{
	for(int i=0;i<=M;i++)
		PS[i][0]=0;
	for(int j=0;j<=N;j++)
		PS[0][j] =0;
	for(int i=1;i<=M;i++)
		for(int j=1;j<=N;j++)
			PS[i][j] = PS[i-1][j]+PS[i][j-1]-PS[i-1][j-1]+A[i-1][j-1];
	int max = -10000;
	for(int a=1;a<=M;a++)
		for(int c=1;c<=N;c++)
		{
			int start = BC(a,c,M);
			int all = BC(a,c,M);
			for(int i=M-1;i>=1;i--)
			{
				if(start<0)
					start = 0;
				start += BC(a,c,i);
				if(start>all)
					all = start;
			}
			if(all>max)
				max = all;
		}
		return max;
}
int main()
{
	int a[M][N] = {-1,-2,-3,-4,-5,-3,4,5,6,-9,-12,44,32,7,-8,-3,22,56,78,-6,-5,2,3,7,-1};	
	cout<< MaxSum(a);
	return 0;
}