POJ-2479 Maximum sum （不连续的最大子段和）

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34398		Accepted: 10661

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

大致题意：给一个数列，求出数列中不想交的两个子段和，要求和最大。

思路：对于每个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，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】从右向左扫描即可。

#include<cstdio>
#include<algorithm>
using namespace std;
int a[50001];
int left[50001];
int right[50001];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        left[0]=a[0];
        for(int i=1;i<n;i++)
        {
            if(left[i-1]<0)
                left[i]=a[i];
            else
                left[i]=left[i-1]+a[i];
        }
        for(int i=1;i<n;i++)
            left[i]=max(left[i],left[i-1]);
        right[n-1]=a[n-1];
        for(int j=n-2;j>=0;j--)
        {
            if(right[j+1]<0)
                right[j]=a[j];
            else
                right[j]=right[j+1]+a[j];
        }
        for(int i=n-2;i>=0;i--)
            right[i]=max(right[i+1],right[i]);
        int res=-100000000;
        for(int i=1;i<n;i++)
        {
            res=max(res,left[i-1]+right[i]);
        }
        printf("%d
",res);
    }
    return 0;
}