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