Pku acm 2479-Maximum Sum

A Dynamic Planning question

Http://acm.pku.edu.cn/JudgeOnline/problem? Id = 2479

 

Maximum Sum

Time limit:1000 ms		Memory limit:65536 K
Total submissions:17327		Accepted:5239

Description

Given a set of N integers: A = {A1, A2,..., an}, we define a function D (A) as below:

The formula is invisible, that is, the maximum value of the sum of the two sub-arrays in array.

 

Your task is to calculate d ().

Input

The input consists of T (<= 30) test cases. The number of test cases (t) is given in the first line of the input.
Each test case contains two lines. the first line is an integer N (2 <= n <= 50000 ). the second line contains N integers: A1, A2 ,..., an. (| ai | <= 10000 ). there is an empty line after each case.

Output

Print exactly one line for each test case. The line shoshould contain the integer d ().

Sample Input

1101 -1 2 2 3 -3 4 -4 5 -5

Sample output

13

Hint

In the sample, we choose {2, 2, 3,-3, 4} and {5}, then we can get the answer.

Huge input, scanf is recommended.

Solution Report: Find the largest sum of array Neutron arrays, which is mentioned in the beauty of programming. However, what is required here is the maximum sum of the two sub-arrays. It is traversed twice. During the first traversal, the greatest sum of the sub-arrays in the array before each number is obtained, during the second traversal, obtain the largest sum of the subarrays in the array after each number, and add the largest sum to the subarrays before it to obtain the final maximum value. The first traversal goes from front to back, and the second traversal goes from back to back. For the first traversal, the maximum sum of the subarray that contains a [I] must be recorded before start [I]: A [I. All [I]: the largest sum of sub-arrays before a [I. The second traversal directly calculates nstart, Nall, maximum value = Nall + ALL [I-1] complexity analysis, O (N) # Include <iostream> <br/> using namespace STD; </P> <p> # define N-500000001 </P> <p> int returnmax (int, int B) <br/>{< br/> return A> B? A: B; <br/>}</P> <p> int main () <br/>{< br/> int casenum; <br/> CIN> casenum; <br/> while (casenum --) <br/> {<br/> int max = N; <br/> int N; <br/> CIN> N; <br/> int * A = new int [N]; <br/> int * Start = new int [N]; <br/> int * All = new int [N]; <br/> int nstart; <br/> int Nall; <br/> for (INT I = 0; I <n; I ++) <br/>{< br/> scanf ("% d", & A [I]); <br/> if (I> 0) <br/> {<br/> Start [I] = returnmax (A [I], start [I-1] + A [I]); <br/> All [I] = returnmax (ALL [I-1], start [I]); <br/>}< br/> else <br/>{< br/> Start [I] = A [I]; <br/> All [I] = A [I]; <br/>}< br/> for (INT I = n-1; I> 0; I --) <br/>{< br/> if (I = N-1) <br/>{< br/> nstart = A [I]; <br/> Nall = A [I]; <br/> max = returnmax (max, Nall + ALL [I-1]); <br/>}< br/> else <br/> {<br/> nstart = returnmax (A [I], nstart + A [I]); <br/> Nall = returnmax (Nall, nstart); <br/> max = returnmax (max, Nall + ALL [I-1]); <br/>}< br/> printf ("% d/N", max); <br/> Delete []; <br/> Delete [] Start; <br/> Delete [] All; <br/>}< br/>}