To the Max POJ - 1050

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

题意:  求二维最大子矩阵的和 . 3层循环枚举出正方形大矩阵的子区域矩阵. 这算是个暴力枚举求子矩阵和最大.

 

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define Swap(a,b)  a ^= b ^= a ^= b
#define pi acos(-1)
#define cl(a,b) memset(a,b,sizeof(a))
using namespace std ;
typedef long long LL;
//const int N = 1e7+10 ;
const int inf = 0x3f3f3f3f;
const int MAX = 105;

int n ;
int a[MAX][MAX] ;
int dp[MAX][MAX] ;
int main()
{
	
	scanf("%d",&n);
	for(int i = 0 ; i>a[i][j] ;
		}
	}
	int maxn = -128 ;
	for(int i = 0 ; ii ){
					a[i][k]+=a[j][k];
				}
				if(b>0){
					b+=a[i][k] ;
				}
				else{
					b = a[i][k] ;
				}
				maxn = max(maxn,b) ;
			}
		}
	}
	cout<