hole may b- separate 轉化 ima first lin con

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

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

Sample Output

15

題解：

看起來很難做的題目，但他讓我們想起 最大字段和，但那個是一維的。

那怎麽轉化二維為1維呢？

當然是枚舉了。

分別枚舉第 i 行到第 j 行 ，這裏的復雜度為O（n^2），然後在用字段和遍歷一遍，所以總的時間復雜度為O（n^3） 考慮到n不大，可行。

有一個問題是怎麽最快求第k列中，第 i 行道第 j 行的和，我們采用前綴和的方式。

由於網上多是行前綴和，這裏貼一個列前綴和的代碼

#include<stdio.h>
#include<cmath>
#include<iostream>
#define INF 0x3f3f3f3f
#define me(a,b) memset(a,b,sizeof(a))
#define N 102
typedef long long ll;
using namespace std;

int n,a[N][N],dp[N][N],t,sum,ans;

int main()
{
      //freopen("input.txt","r",stdin);
      cin>>n;
      for(int i=1;i<=n;i++)
      {
            for(int j=1;j<=n;j++)
            {
                  cin>>t;
                  a[i][j]=a[i][j-1]+t;//記錄前綴和
            }
      }
      for(int i=1;i<=n;i++)
      {
            for(int j=i;j<=n;j++) //枚舉，第i列道第j列
            {
                  sum=0;
                  for(int k=1;k<=n;k++)//最大子序列遍歷一遍
                  {
                        int t=a[k][j]-a[k][i-1];//t代表： 第k行中，第 i 列到第 j 列的和
                        sum+=t;
                        sum=sum<0?0:sum; //<0則置為0
                        if(sum>ans)
                              ans=sum;
                  }
            }
      }
      cout<<ans;

}

To the Max 二維dp（一維的變形）