Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 51076 Accepted: 27026
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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
题目大意：输入一个N*N的整数矩阵，包含正负数，从该矩阵中寻找一个子矩阵使其数之和最大，子矩阵最小可为1*1。
二维空间寻找子矩阵较复杂，可以先降到一维先考虑如何求得一维数组子段和最大。
求最大子段和：
int max_array(int a[],int n)
{
int sum=0,max=-10000;
for(int i=0;i<n;i++)
{
if(sum>0) //如果到此为止和为正的话继续加入下一元素比较
sum+=a[i];
else //如果和为零或负值则进行剪枝，直接以下一元素作为字段头再开始
sum=a[i];
if(sum>max) //记录每次的最大值并不断更新
max=sum;
}
return max;
}
接下来从最大子段为基础考虑最大子阵：
int max_rectangle(int n,int b[][101])
{
int temp[101];
int max=-10000;
for(int i=0;i<n;i++)
{
memset(temp,0,sizeof(temp)); //注意清空重置！
for(int j=i;j<n;j++)
{
for(int k=0;k<n;k++)
temp[k]+=b[j][k]; //将二维的数组相加压缩成一维的，找最大子段和
if(max_array(k,temp)>max)
max=max_array(k,temp);
}
}
return max;
}