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[HDU 1081] To The Max [Java] [Runtime : 358MS]

2017-09-27
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09/27

1. 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.

2. 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].

3. Output

Output the sum of the maximal sub-rectangle.

4. Example

input
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
output
15

5. Code

import java.util.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String args[])
    {
        Scanner in = new Scanner(System.in);

        while(in.hasNextInt())
        {
            int n = in.nextInt();
            int[][] matrix = new int[n][n];

            for(int i = 0; i < n; i++ ){
                for(int j = 0; j < n; j++ ){
                    matrix[i][j] = in.nextInt();
                }
            }

            int result = Integer.MIN_VALUE;
            int[] dp = new int[n];
            for(int i = 0; i< n; i++){
                Arrays.fill(dp, 0);
                for(int j = i; j< n; j++){
                    for(int k = 0; k< n; k++){
                        dp[k] +=matrix[j][k];
                    }
                    result = Math.max(result, maxSubArraySum(dp));
                }
            }
            System.out.println(result);
        }
    }

    private static int maxSubArraySum(int[] array){
        int pre = 0;
        int sum = Integer.MIN_VALUE;
        for(int i = 0; i < array.length; i++ ){
            if(pre < 0){
                pre = array[i];
            }else{
                pre += array[i];
            }
            if(sum < pre){
                sum = pre;
            }
        }
        return sum;
    }
}
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