一、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.
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].
(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.
二、题解
这个题目怎么说呢，说难不难，说不难呢也花了我一天时间。刚开始的时候没什么思路，看到有讨论说是DP。但看了想了好久，也没找到状态转换方程，因为有不定长的X，Y的情况。在纠结了片刻后，我决定暴力解决，采用最简单的枚举方法，写出来我自己的惊呆了，O(N^6)。天啊，想了一下一定过不了。果然，TLE了。又想了几刻钟，看了下讨论，发现了一种不错的思路，把二维数组压缩成一维数组，然后再求最大子段和。这个压缩过程其实就是把第i+1行依次加到第i（0<= i <=n-1）行然后求最大子段和，记录最大值就OK了。
三、java代码
二、题解
这个题目怎么说呢，说难不难，说不难呢也花了我一天时间。刚开始的时候没什么思路，看到有讨论说是DP。但看了想了好久，也没找到状态转换方程，因为有不定长的X，Y的情况。在纠结了片刻后，我决定暴力解决，采用最简单的枚举方法，写出来我自己的惊呆了，O(N^6)。天啊，想了一下一定过不了。果然，TLE了。又想了几刻钟，看了下讨论，发现了一种不错的思路，把二维数组压缩成一维数组，然后再求最大子段和。这个压缩过程其实就是把第i+1行依次加到第i（0<= i <=n-1）行然后求最大子段和，记录最大值就OK了。
三、java代码
import java.util.Scanner;
public class Main {
static int n;
static int MaxSub (int a[], int N){
int max, i;
int[] dp=new int [n];
max = dp[0] = a[0];
for (i=1; i<N; i++){
if (dp[i-1] > 0)
dp[i] = dp[i-1] + a[i];
else
dp[i] = a[i];
if (dp[i] > max)
max = dp[i];
}
return max;
}
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
int i, j, k, Max, m;
n=cin.nextInt();
int[][] a =new int [n][n];
for (i=0; i<n; i++) {
for (j=0; j<n; j++){
a[i][j]=cin.nextInt();
}
}
Max = Integer.MIN_VALUE;
for (i=0; i<n; i++){
m = MaxSub(a[i], n);
if (m > Max)
Max = m;
for (j=i+1; j<n; j++){
for (k=0; k<n; k++){
a[i][k] += a[j][k];
}
m = MaxSub(a[i], n);
if (m > Max)
Max = m;
}
}
System.out.print(Max);
}
}
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