编程之美子数组之和的最大值

通过确定矩阵区域的上下边界，将二维问题转化为一维问题。时间复杂度转化为O（N*M*min（N,M））具体分析见编程之美。

代码如下：

#include <stdio.h>
#include <limits.h>

int BC(int** A, int, int, int);

int maxSum(int** A, int n, int m)
{
	int maximum = INT_MIN;
	for (int i = 0; i < n; i++)
	{
		for (int j = i; j < n; j++)
		{
			int nStart = BC(A, i, j, m-1);
			int nAll = BC(A, i, j, m-1);

			for (int k = m-2; k >= 0; k--)
			{			
				if (nStart < 0)
				{
					nStart = 0;
				}
				nStart += BC(A, i, j, k);
				if (nStart > nAll)
				{
					nAll = nStart;
				}
			}
			if (nAll > maximum)
			{
				maximum = nAll;
			}
		}
	}
	return maximum;
}

int BC(int** A, int i, int j, int m)
{
	int sum = 0;
	for (int k = i; k <= j; k++)
	{
		sum += A[k][m];
	}
	return sum;
}

int main()
{
	int n, m, sum;
	int** A;
	printf("please input n and m:\n");
	while (scanf("%d %d",&n,&m)!=EOF)
	{
		A = new int*[n];
		for (int i = 0; i < n; i++)
		{
			A[i] = new int[m];
		}
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
			{
				scanf("%d",&A[i][j]);
			}
		}
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
			{
				printf("%d ",A[i][j]);
			}
			printf("\n");
		}
		sum = maxSum(A, n, m);
		printf("the max sum is %d",sum);
		for (int i = 0; i < n; i++)
		{
			delete[]A[i];
		}
		delete[]A;
		printf("please input n and m:\n");
	}
}