poj----Maximum sum(poj 2479)

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30704		Accepted: 9408

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

1

10
1 -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

题意：

给一个数列，求出数列中不相交的两个字段和，要求和最大。

思路:

对每一个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，再加起来求最大的一个就行了。[0~i-1]的最大子段和从左向右扫描，【i~n-1】从右想左扫描

即可，时间复杂度O(n).

代码：

 #include<iostream>
 #include<cstdio>
 #define maxn 50001
 #include<algorithm>
 using namespace std;
 int  a[maxn];
 int left[maxn];
 int right[maxn];
 int max(int a,int b)
 {
     return a>b?a:b;
 }
 int main()
 {
     int t,i;
     scanf("%d",&t);
     while(t--)
     {
         int n;
       scanf("%d",&n);
       for(i=;i<n;i++)
           scanf("%d",&a[i]);
       //此时::left【i】为包涵i最大字段和
          ::left[]=a[];
       for( i=; i<n;i++)
           if(::left[i-]<)
               ::left[i]=a[i];
           else
               ::left[i]=::left[i-]+a[i];
      //此时left[i]为i左边最大字段和
         for(i=; i<n;i++)
             ::left[i]=max(::left[i],::left[i-]);
             ::right[n-]=a[n-];
         for(i=n-;i>=;i--)
         {
             if(::right[i+]<)
                 ::right[i]=a[i];
             else
                 ::right[i]=::right[i+]+a[i];
         }
         for(i=n-;i>=;i--)
             ::right[i]=max(::right[i+],::right[i]);
         int res=-;
         for(i=;i<n;i++)
         {
             res=max(res,::left[i-]+::right[i]);
         }
         printf("%d\n",res);
     }
   return ;
 }