（线性dp 最大子段和 最大子矩阵和）POJ1050 To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54338		Accepted: 28752

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

就是最大字段和的升级版，
从：http://www.cnblogs.com/fll/archive/2008/05/17/1201543.html 可知：

假设最大子矩阵的结果为从第r行到k行、从第i列到j列的子矩阵，如下所示(ari表示a[r][i],假设数组下标从1开始)：
  | a11 …… a1i ……a1j ……a1n |
  | a21 …… a2i ……a2j ……a2n |
  |  .     .     .    .    .     .    .   |
  |  .     .     .    .    .     .    .   |
  | ar1 …… ari ……arj ……arn |
  |  .     .     .    .    .     .    .   |
  |  .     .     .    .    .     .    .   |
  | ak1 …… aki ……akj ……akn |
  |  .     .     .    .    .     .    .   |
  | an1 …… ani ……anj ……ann |

那么我们将从第r行到第k行的每一行中相同列的加起来，可以得到一个一维数组如下：
 (ar1+……+ak1, ar2+……+ak2, ……,arn+……+akn)
 由此我们可以看出最后所求的就是此一维数组的最大子断和问题，到此我们已经将问题转化为上面的已经解决了的问题了。

就是先让i从0到n遍历，然后j从i到n遍历，最后在第j行中k从0到n遍历，用一个数组分别保存每个列的各行的数字之和，就可以化为最大连续和（降维）。

复杂度为：O（n^3）

C++代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = ;
int d[maxn][maxn];
int s[maxn];
int INF = -0x3f3f3f3f;

int MaxArray(int a[],int n){
    int m = INF;
    int tmp = -;
    for(int i = ; i < n; i++){
        if(tmp > )
            tmp += a[i];
        else
            tmp = a[i];
        if(tmp > m)
            m = tmp;
    }
    return m;
}

int main(){
    int n;
    scanf("%d",&n);
    for(int i = ; i < n; i++){
        for(int j = ; j < n; j++){
            scanf("%d",&d[i][j]);
        }
    }
    int ans = INF,tmp;
    for(int i = ; i < n; i++){
        memset(s,,sizeof(s));
        for(int j = i; j < n; j++){
            for(int k = ; k < n; k++){
                s[k] += d[j][k];
            }
            tmp = MaxArray(s,n);
            if(tmp > ans)
                ans = tmp;
        }
    }
    printf("%d\n",ans);
    return ;
}