poj2479

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27256		Accepted: 8329

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

 

DP: 最大子段和

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int dp1[50010],dp2[50010];
int a[50010];
int n,t;
int maxD;

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        //dp for 最大子序列和
        dp1[0]=-10000000;
        for(int i=1; i<=n; i++)
        {
            if(dp1[i-1]<0)
                dp1[i]=a[i];
            else
                dp1[i]=dp1[i-1]+a[i];
        }
        for(int i=1; i<=n; i++)
        {
            if(dp1[i-1]>dp1[i])
            {
                dp1[i]=dp1[i-1];
            }
        }

        //reverse dp
        dp2[n+1]=-10000000;
        for(int i=n; i>=1; i--)
        {
            if(dp2[i+1]<0)
                dp2[i]=a[i];
            else
                dp2[i]=dp2[i+1]+a[i];
        }
        for(int i=n; i>=1; i--)
        {
            if(dp2[i+1]>dp2[i])
                dp2[i]=dp2[i+1];
        }

        //cal
        maxD=-10000000;
        for(int i=1; i<=n-1; i++)
        {
            if(dp1[i]+dp2[i+1] > maxD)
                maxD=dp1[i]+dp2[i+1];
        }

        printf("%d\n",maxD);
    }
    return 0;
}