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

     return ;
 }