HDU 1081 to the Max (dp+ two-dimensional one-dimensional)

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1081

to the Max

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 8839 Accepted Submission (s): 4281

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:

7 |
-4 1
-1 8

and has a sum of 15.

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

Sample Output15 topic: Find a matrix inside a number and the largest matrix. The concept of the topic: the two-dimensional one-dimensional, converted into 1003 can. First, one column and one column, and this translates into one dimension. For example, I first calculate the first line to get dp[0]=0dp[1]=-2,dp[2]=-7,dp[3]=0, these are called and, and then in the 1003 method. Then the second line, will be updated, dp[0]=9,dp[1]=0,dp[2]=-13,dp[3]=2; these are still and, this is converted into a one-dimensional array, with 1003 of the method to solve, and so on ~ ~ ~ See the code.

1#include <iostream>2#include <cstdio>3#include <cstring>4 5 using namespacestd;6 7 intdp[ the];8 intnum[ the][ the];9 Ten intMain () One { A     intN; -      while(~SCANF ("%d",&N)) -     { the          for(intI=0; i<n; i++) -              for(intj=0; j<n; J + +) -scanf"%d",&num[i][j]); -         intans=-99999; +          for(intI=0; i<n; i++) -         { +Memset (DP,0,sizeof(DP)); A              for(intJ=i; j<n; J + +) at             { -                 intmax=-1; -                  for(intk=0; k<n; k++) -                 { -DP[K]+=NUM[J][K];//calculate all the DP and -                 } in                  for(intk=0; k<n; k++)//1003 approach, code similar -                 { to                     if(max+dp[k]<Dp[k]) +max=Dp[k]; -                     Else themax=max+Dp[k]; *                     if(Ans<max)//constantly updating the maximum value $                     {Panax Notoginsengans=Max; -                         //cout<<j<< "" <<k<<endl; the                     } +                 } A             } the         } +printf ("%d\n", ans); -     } $     return 0; $}

HDU 1081 to the Max (dp+ two-dimensional one-dimensional)