POJ2479(dp)

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 39089		Accepted: 12221

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.

分别求出两端开始的最大子段和，然后枚举左右两段的分界，找出最大值。

//2016.8.21
#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N = 50005;
const int inf = 0x3f3f3f3f;
int a[N], dpl[N], dpr[N];//dpl[i]表示从左往右到第i位的最大子段和,dpr[i]表示从右往左到第i位的最大子段和

int main()
{
    int T, n;
    cin>>T;
    while(T--)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
        }
        memset(dpl, 0, sizeof(dpl));
        memset(dpr, 0, sizeof(dpr));
//从左往右扫        
//*************************************************        
        dpl[0] = a[0];
        for(int i = 1; i < n; i++)
            if(dpl[i-1]>0) dpl[i] = dpl[i-1]+a[i];
            else dpl[i] = a[i];
        for(int i = 1; i < n; i++)
            if(dpl[i]<dpl[i-1])
                  dpl[i] = dpl[i-1];
//从右往左扫        
//*************************************************        
        dpr[n-1] = a[n-1];
        for(int i = n-2; i>=0; i--)
            if(dpr[i+1]>0) dpr[i] = dpr[i+1]+a[i];
            else dpr[i] = a[i];
        for(int i = n-2; i>=0; i--)
            if(dpr[i]<dpr[i+1])
                  dpr[i] = dpr[i+1];
//*************************************************        
        int ans = -inf;
        for(int i = 0; i < n-1; i++)
        {
            ans = max(ans, dpl[i]+dpr[i+1]);
        }
        cout<<ans<<endl;
    }

    return 0;
}