HDU 1081 To the Max 最大子矩阵（动态规划求最大连续子序列和）

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

Sample Output

该题和之前做过的求最大连续子序列和有着共通之处，不同的地方是本问题求的是一个最大连续子矩阵，维度上变成了二维，那么我们能不能考虑压缩二维空间变为一维空间呢？

其实这是可以的。可以先选中矩阵的几行，逐列相加变为一个一维数组。（如果是一行的情况，则直接使用序列的最大连续段和方法）
多行变为一行时：例如
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
当i=0， j=2时，则选择0，1，2行，逐列相加压缩为一行a，再从a中寻找最长连续子序列和
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
a: 5 1 -17 3
该问题中i和j需要遍历所有的情况，压缩所有情况为一个一维数组！！！

import java.util.*;

public class test {

  public static void main(String[] args) {

    int[][] a = new int[100][100];

    int[] b = new int[100];

    int n;

    Scanner in = new Scanner(System.in);

    n = in.nextInt();

    for (int i = 0; i < n; i++) {

      for (int j = 0; j < n; j++) {

        a[i][j] = in.nextInt();

      }

    }

    int ans = 0;

    for (int i = 0; i < n; i++) {

      for (int j = i; j < n; j++) {

        for (int k = 0; k < n; k++) {

          b[k] = 0;

          for (int l = i; l <= j; l++) {

            b[k] += a[l][k];//合并i到j行

          }

        }

        // 动态规划

        int sum = 0;//当前和

        int max = 0;//最大和

        //dp[i] = max(dp[i], dp[i - 1] + a[i]);

        for (int k = 0; k < n; k++) {

          sum += b[k];// 含有第k个元素的最大连续子段和

          if (sum > max) {

            max = sum;

          }

          if (sum < 0) {

            sum = 0;

          }

        }

        if (max > ans) {//更新ans

          ans = max;

        }

      }

    }

    System.out.println(ans);

  }

}

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