【noiOJ】p1481

1481:Maximum sum

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总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

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

                     t1     t2 
         d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n }
                    i=s1   j=s2

Your task is to calculate d(A).

输入

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.

输出

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

样例输入

1

10
1 -1 2 2 3 -3 4 -4 5 -5

样例输出

13

提示

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

Huge input,scanf is recommended.

来源

POJ Contest,Author:Mathematica@ZSU

source:最大双子序列和，枚举i，分别统计i左边最大子序列和右边最大子序列。（改变一下代码风格）

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
int n,datanum,maxn;
int a[50000+10],leftmax[50000+10],rightmax[50000+10];
int main()
{
    int w,i;
    scanf("%d",&datanum);
    for (w=1;w<=datanum;w++)
    {
        memset(a,0,sizeof(0));
        memset(leftmax,0,sizeof(leftmax));
        memset(rightmax,0,sizeof(rightmax));
        scanf("%d",&n);
        for (i=1;i<=n;i++)
            scanf("%d",&a[i]);
        leftmax[1]=a[1];
        for (i=2;i<=n;i++)
            if (leftmax[i-1]>0)        leftmax[i]=leftmax[i-1]+a[i];
            else    leftmax[i]=a[i];
        for (i=2;i<=n;i++)
            if (leftmax[i]<leftmax[i-1])    leftmax[i]=leftmax[i-1];
        rightmax[n]=a[n];
        for (i=n-1;i>=1;i--)
            if (rightmax[i+1]>0)    rightmax[i]=rightmax[i+1]+a[i];
            else    rightmax[i]=a[i];
        for (i=n-1;i>=1;i--)
            if (rightmax[i]<rightmax[i+1])
                rightmax[i]=rightmax[i+1];    
        maxn=-99999999;
        for (i=1;i<n;i++)
            maxn=max(maxn,leftmax[i]+rightmax[i+1]);
        printf("%d\n",maxn);
    }
    return 0;
}