To the Max

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15 总结      虽然做了dp有一段时间了，但感觉还是一脸懵逼，看到这道题时完全没有思路，然后又是各种翻博客，看了半天才明白应该用什么思路，或许我就是一个 见得多才做的了题的人吧，以前弄奥数也是，很少有自己一次性做出来的新题，我要总结过这些思路后才想到怎么做。总之，在这个题我也学到了许多， 如何将二维转化为一维，如何求最长子串（请原谅我忘记了o(╯□╰)o）。

#include
     
      #include
      
       #include
       
        using namespace std;#define inf 0x3f3f3f3f;int mat[105][105];int temp[105];int n;int main(){    int max;    cin>>n;    for(int i=1;i<=n;i++){        for(int j=1;j<=n;j++){            scanf("%d",&mat[i][j]);        }    }    max=-inf;    for(int i=1;i
        
         max1) max1=sum;            }            if(max