POJ 1050 DP

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 44351 Accepted: 23510

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

#include<cstdio>
#include<iostream>
using namespace std;
int dp[102][102]={0};
int main()
{
int n=0;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
int x=0;
scanf("%d",&x);
dp[i][j]=dp[i][j-1]+x;
}
int Max=0;
   for(int i=0;i<=n-1;i++)
    for(int j=i+1;j<=n;j++)
    {
    int max=0;
    int sum=0;
    for(int k=1;k<=n;k++)
    {
     sum+=dp[k][j]-dp[k][i];
     if(sum>max)
     {
     max=sum;
 }
 if(sum<0)
 {
  sum=0;
 }
}
if(max>Max) Max=max;
}
printf("%d\n",Max);
}
return 0;
}