To the Max 二维dp（一维的变形）

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

题解：

看起来很难做的题目，但他让我们想起 最大字段和，但那个是一维的。

那怎么转化二维为1维呢？

当然是枚举了。

分别枚举第 i 行到第 j 行 ，这里的复杂度为O（n^2），然后在用字段和遍历一遍，所以总的时间复杂度为O（n^3） 考虑到n不大，可行。

有一个问题是怎么最快求第k列中，第 i 行道第 j 行的和，我们采用前缀和的方式。

由于网上多是行前缀和，这里贴一个列前缀和的代码

#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=;i<=n;i++)
      {
            for(int j=;j<=n;j++)
            {
                  cin>>t;
                  a[i][j]=a[i][j-]+t;//记录前缀和
            }
      }
      for(int i=;i<=n;i++)
      {
            for(int j=i;j<=n;j++) //枚举，第i列道第j列
            {
                  sum=;
                  for(int k=;k<=n;k++)//最大子序列遍历一遍
                  {
                        int t=a[k][j]-a[k][i-];//t代表： 第k行中，第 i 列到第 j 列的和
                        sum+=t;
                        sum=sum<?:sum; //<0则置为0
                        if(sum>ans)
                              ans=sum;
                  }
            }
      }
      cout<<ans;

}