poj 2479 dp求分段最大和

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 38079		Accepted: 11904

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.

题目大意：将区间分成了两部分，让你求这两部分的和最大是多少。

思路分析：前面做过求最大连续序列和的题目，是用dp做的，状态转移方程为dp[i]=max[dp[i-1]+a[i],a[i])

这道题与那道题很像，唯一的区别就是区间变成了两部分，现在要求和最大，dp[i]的意思是以i结尾的子串的最大

和，现在可以把分成求两个dp，l[i]代表i之前的最大和,r[i]代表i之后的最大和，题目要求的不就是求l[i-1]+r[i]的

最大值么，现在问题就变成了如何求了l[i]和r[i],状态转移方程很容易确定，l[i]=max(l[i-1],t1[i])，t1[i]指的是
以i个数结尾的最大子串和，这样问题经过几步分解就变成了已经解决过的问题了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;
const int maxn=50000+100;
int a[maxn],t1[maxn],t2[maxn],l[maxn],r[maxn];
const int inf=-1000001;
int main()
{
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int n;
            scanf("%d",&n);
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
            }
            memset(t1,0,sizeof(t1));
            memset(t2,0,sizeof(t2));
            for(int i=1;i<=n;i++)
            {
                t1[i]=max(t1[i-1]+a[i],a[i]);
            }
            for(int i=n;i>=1;i--)
            {
                t2[i]=max(t2[i+1]+a[i],a[i]);
            }
            l[1]=a[1];
            for(int i=2;i<=n;i++)
            {
                l[i]=max(l[i-1],t1[i]);
            }
           r[n]=a[n];
           for(int i=n-1;i>=1;i--)
           {
               r[i]=max(r[i+1],t2[i]);
           }
           int ma=inf;
           for(int i=2;i<=n;i++)
           {
               int t=l[i-1]+r[i];
               ma=max(ma,t);
           }
           cout<<ma<<endl;
        }
     return 0;
}