Interval DP brute force enumeration + Dynamic Planning hdu1081

F-Maximum child rectangle
Time Limit: 1000 ms memory limit: 10000kb 64bit Io format: % i64d & % i64u
Submit

Status
Description
Given a two-dimenstmarray 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-dimen=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

<Span style = "color: # 3333ff;">/* revoke Author: grant yuan time: 2014.7.19 Source: hdu1081 algorithm: Brute Force enumeration + Dynamic Planning explain: brute-force enumeration of sum [I] of each element from row K1 to row K2, then sum [I] is summed up in the largest subsequence _______________________________________________________________________________________________________________________________________________________ Lude <iostream> # include <cstdio> # include <cstring> # include <cstdlib> # include <functional> # include <algorithm> using namespace STD; int A [105] [105]; int L; int sum [105]; int DP [105]; int main () {While (~ Scanf ("% d", & L) {for (INT I = 0; I <L; I ++) for (Int J = 0; j <L; j ++) scanf ("% d", & A [I] [J]); int ans =-9999999, ans1; For (INT k1 = 0; k1 <L; k1 ++) for (INT k2 = 0; K2 <L; K2 ++) {memset (DP, 0, sizeof (DP); memset (sum, 0, sizeof (SUM); For (INT I = 0; I <L; I ++) {for (Int J = k1; j <= k2; j ++) {sum [I] + = A [I] [J] ;}for (INT m = 0; m <L; m ++) {DP [M + 1] = max (DP [m] + sum [m], sum [m]);} ans1 = DP [1]; for (INT d = 1; D <= L; D ++) if (DP [d]> ans1) ans1 = DP [d]; If (ans1> ans) ans = ans1 ;}} printf ("% d \ n", ANS) ;}}</span>