POJ 1050 To the Max (动规)

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To the Max

Time Limit: 1000MS
Memory Limit: 10000KTotal Submissions: 40302
Accepted: 21329

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

Source

Greater New York 2001

在一维情况下最大连续子段和的求法是从左到有顺序扫描数据，以0为边界，当累加和小于0时则重置为0.动态规划的状态转移方程为

s=max{s_i-1+ai,ai}，该方程和前面的描述是等价的。本题是对一维最大子段和的扩展，思路是从上到下找出所有的连续行(如第i行到第j行)，然后计算每列从第i行到第j行的和，之后对这n个列的和进行一维最大子段和的计算，并找出最大的值。

最大子矩阵，首先一行数列很简单求最大的子和，我们要把矩阵转化成一行数列，就是从上向下在输入的时候取和，map[i][j]表示在J列从上向下的数和，这样就把一列转化成了一个点，再用双重，循环，任意i行j列开始的一排数的最大和，就是最终的最大和

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。

代码：

#include <iostream>using namespace std;#define M 110int main(){int a[M][M]={0},c[M][M]={0}; //a[M][M]用来存数，c[M][M]就是存这行到这一列的和。int i,j,n,max=0,sum,k;    scanf("%d",&n);for(i=0;i<n;i++)  for(j=1;j<=n;j++)  {        scanf("%d",&a[i][j-1]);        c[i][j]=c[i][j-1]+a[i][j-1];   //一行的下一个数和上面所有数的和。  }   for(i=0;i<n;i++)  for(j=i;j<=n;j++)  {  sum=0;  for(k=0;k<n;k++)  {  sum+=c[k][j]-c[k][i];  if(sum<0) sum=0;  //小于0就相当于不用取了，直接去掉   else if(sum>max) max=sum;  }  }  printf("%d\n",max);  return 0;}

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