To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6448    Accepted Submission(s): 3088

Problem Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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 
     
 
    
 
    

        
    
 
    
 
    

      Source 
    
 
    

      Greater New York 2001 
     
 
     
 
     
 
     

 解题思路：本题为动态规划题目，其原题型为（http://blog.csdn.net/lsh670660992/article/details/9498821）  ，原题是求一维数组的最大子串和，本题要求二维数组的最大子二维数组和。看似联系不大。但是，要是把数组中的相关连续列（不中断）合并到一个一维数组（模板数组），求出数组的最大子串和，就是原二维数组的一个子串和了，只要把所有的相关列枚举，求出其模板一维数组的最大子串和，再求出所有子串和中得最大值，就可以得到二维数组的最大子串和了。 
     
 
     
 
     
 
     
 
     
#include #include int n,a[102]; int fmax() {     int sum=0;     int max1=-0xfffffff;     for(int i=0;imax1)             max1=sum;         if(sum<0)             sum=0;     }     return max1; } int main() {     int i,j,k;     int s[102][102];     while(scanf("%d",&n)!=EOF)     {         int smax=-0xfffffff;         for(i=0;ix?smax:x;             }         printf("%d\n",smax);     }     return 0; }