HDU 1081 To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14951    Accepted Submission(s): 6983

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

 DP

#include<iostream>
//#include<fstream>
#include<memory.h>
using namespace std;
int main()
{
    //ifstream in("data.txt");
    int n,i,j,k;
    int mat[102][102]={0};
    int a;
    while(cin>>n)
    {
        memset(mat,0,sizeof(mat));
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                cin>>a;
                mat[i][j]=mat[i][j-1]+a;
            }
        int max=-128,sum;
        for(i=1;i<=n;i++)
            for(j=i;j<=n;j++)
            {
                sum=0;
                for(k=1;k<=n;k++)
                {
                    sum+=mat[k][j]-mat[k][i-1];
                    if(sum>max)
                        max=sum;
                    if(sum<0)
                        sum=0;
                }
            }
        cout<<max<<endl;
    }
    return 0;
}