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

解題思路：這個和一維陣列擷取一個欄位得到的最值是一一樣的，只是一維變成二維的，所以要對 資料進行處理，使得像一維資料那樣處理，用一個三維的陣列來儲存一個列裡，從i行到j行的和，然 後接著就是像一維陣列一樣從1~n列，只是這個不是一個數組而已，加了很多陣列，這樣就可以取他們 中最大的最大即可。（動態規劃） 
程式碼如下： 

Code:

1. #include<iostream>
2. usingnamespace std;  
3. constint Max(101);   
4. int sum[Max][Max][Max];     //用來儲存在一個列裡，從i行到j行的和
5. int main()  
6. {     
7.     int n;  
8.     while(cin>>n)  
9.     {     
10.         int data[Max][Max];  
11.         for(int i=1;i<=n;i++)  
12.         {  
13.             for(int j=1;j<=n;j++)  
14.             {  
15.                 cin>>data[i][j];  
16.                 sum[i][i][j]=data[i][j];        //先初始化為自己那一格的值
17.             }  
18.         }  
19.         for(int i=1;i<=n;i++)  
20.         {  
21.             for(int j=i+1;j<=n;j++)  
22.             {  
23.                 for(int k=1;k<=n;k++)  
24.                 {  
25.                     sum[i][j][k]=sum[i][j-1][k]+data[j][k];       //計算出在k列從i行到j行的和
26.                 }  
27.             }  
28.         }  
29.         int MaxSum=-9999;                     //先讓初始化他們的最大值
30.         for(int i=1;i<=n;i++)  
31.         {  
32.             for(int j=i;j<=n;j++)  
33.             {  
34.                 int temp=0;  
35.                 for(int k=1;k<=n;k++)                //和一維陣列的得最大的段和是一樣的操作
36.                 {  
37.                     if(temp>=0)  
38.                         temp+=sum[i][j][k];  
39.                     else
40.                         temp=sum[i][j][k];  
41.                     if(temp>MaxSum)  
42.                         MaxSum=temp;  
43.                 }  
44.             }  
45.         }  
46.         cout<<MaxSum<<endl;  
47.     }  
48.     return 0;     
49. }