HDU 1081 to the Max basic DP Good question

　　　　　　　　　　　　　　　　　　to the Max

problem DescriptionGiven A two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 X 1 or greater located within the whole array. The sum of a rectangle is the sum of the "the elements in" that rectangle. The problem the sub-rectangle with the largest sum are 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. 

InputThe input consists of an n x n array of integers. The input begins with a single positive integer N in a line by itself, indicating the size of the square two-dimensional a Rray. This was followed by N 2 integers separated by whitespace (spaces and newlines). These is the N 2 integers of the array, presented in Row-major order. That's, all numbers on the first row, left-to-right, then all numbers-second row, left-to-right, etc. N may be as large as 100. The numbers in the array would be in the range [ -127,127]. 

OutputOutput The sum of the maximal sub-rectangle. 

Sample Input40-2-7 09 2-6 2-4 1-4 1-18 0-2 

Sample Output theis to give a n*n matrix and find its maximum sub-matrix .we've all done the maximal subsequence of a sequence. The complexity of O (n). This problem is the problem of transformation. violence enumerates the sum of the K1 rows to each element of each column of row K2 and Sum[i], and then the maximum subsequence summation for Sum[i] and keep updating the answers .

1#include <cstdio>2#include <cstring>3#include <algorithm>4 5 using namespacestd;6 7 Const intmaxn= the;8 9 intA[MAXN][MAXN];Ten intSUM[MAXN]; One intDP[MAXN]; A  - intMain () - { the     intN; -      while(SCANF ("%d", &n)! =EOF) -     { -          for(intI=1; i<=n;i++) +              for(intj=1; j<=n;j++) -scanf"%d",&a[i][j]); +  A         intans=-1000000; at  -          for(intI=1; i<=n;i++) -         { -Memset (DP,0,sizeof(DP)); -memset (SUM,0,sizeof(sum)); -              for(intj=i;j<=n;j++) in             { -                  for(intk=1; k<=n;k++) to                 { +sum[k]+=A[j][k]; -                 } the  *                  for(intk=1; k<=n;k++) $                 {Panax NotoginsengDp[k]=max (dp[k-1],0)+Sum[k]; -                     if(dp[k]>ans) theans=Dp[k]; +                 } A  the             } +         } -  $printf"%d\n", ans); $     } -  -     return 0; the}

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HDU 1081 to the Max basic DP Good question