POJ2479——Maximum sum

Maximum sum

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 33107 Accepted: 10238

Description

Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:

Your task is to calculate d(A).

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 should contain the integer d(A).

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.

Source

POJ Contest,Author:Mathematica@ZSU

分析：简单DP。

对每个i来说，求出[0~i-1]的最大子段和 以及 [i~n-1]的最大子段和，加起来，之后遍历i寻找到最大的和即为答案。

[0~i-1]的最大子段和从左向右，[i~n-1]的最大子段和从右向左。

时间复杂度O(n)。

//Maximum sum//cstdio是将stdio.h的内容用C++头文件的形式表示出来#include<cstdio>#include<algorithm>using namespace std;int a[50001];int left[50001];int right[50001];int main(){int ncase;scanf("%d",&ncase);while(ncase--){int n;scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d",&a[i]);//此时left[i]为包含i最大子段和left[0] = a[0]; // !!!for(int i=1;i<n;i++){if(left[i-1] < 0)left[i] = a[i];elseleft[i] = left[i-1] + a[i];}//此时left[i]为i左边最大子段和for(int i=1;i<n;i++)left[i] = max(left[i], left[i-1]);//right[n-1] = a[n-1];for(int i=n-2;i>=0;i--){if(right[i+1] < 0)right[i] = a[i];elseright[i] = right[i+1] + a[i];}for(int i=n-2;i>=0;i--)right[i] = max(right[i], right[i+1]);//int ans = -10000;for(int i=1;i<n;i++)ans = max(ans, left[i-1]+right[i]);printf("%d\n",ans);}return 0;}