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ZOJ 1074 To the Max (DP)
2015-07-20 17:55:22 】 浏览:5056
Tags:ZOJ 1074 the Max

Problem

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.

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.


Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15


水,直接水过。

#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        using namespace std; int a[110][110]; int n,x; int main() { int sum,maxn; while(~scanf("%d",&n)) { memset(a,0,sizeof(a)); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { scanf("%d",&x); a[i][j]=a[i-1][j]+x; } } maxn=0; for(int i=1;i<=n;i++) { for(int j=i;j<=n;j++) { sum=0; for(int k=1;k<=n;k++) { int t=a[j][k]-a[i-1][k]; sum+=t; // cout<<"fuck "<
       
        maxn) maxn=sum; } } } printf("%d\n",maxn); } return 0; } 
       
      
     
    
   
  


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