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=0;i<n;i++)
          scanf("%d",&a[i]);
      //此时::left【i】为包涵i最大字段和
         ::left[0]=a[0];
      for( i=1; i<n;i++)
          if(::left[i-1]<0)
              ::left[i]=a[i];
          else
              ::left[i]=::left[i-1]+a[i];
     //此时left[i]为i左边最大字段和
        for(i=1; i<n;i++)
            ::left[i]=max(::left[i],::left[i-1]);
            ::right[n-1]=a[n-1];
        for(i=n-2;i>=0;i--)
        {
            if(::right[i+1]<0)
                ::right[i]=a[i];
            else
                ::right[i]=::right[i+1]+a[i];
        }
        for(i=n-2;i>=0;i--)
            ::right[i]=max(::right[i+1],::right[i]);
        int res=-100000000;
        for(i=1;i<n;i++)
        {
            res=max(res,::left[i-1]+::right[i]);
        }
        printf("%d
",res);
    }
  return 0;
}