1481:Maximum sum

题目链接http://noi.openjudge.cn/ch0206/1481/

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.

这道题目的基础是最大子段和。之前学习过有关的知识，自以为可以很轻松的解决这个问题，直到亲手做了一遍才发现不是那么回事。这道题目在最大子段和的基础上强化了，要求两个不想交的子段的和的最大值。这个问题与另一道问题就很类似 here

核心的思路就是先求出从前到后以i为结尾的最大子段和，和到i位置的子段和的最大值。然后从后往前进行相同的过程。最后就是求解和的最大值。

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#define DEBUG(x) cout<<#x<<" = "<<x<<endl
using namespace std;
const int MAXN=5e4+10;;
int N;
int num[MAXN];
int sumf[MAXN];
int sumr[MAXN];
int dpf[MAXN];
int dpr[MAXN];
int main()
{
//    freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&N);
        for(int i=1;i<=N ;i++ ){
            scanf("%d",&num[i]);
        }
        sumf[1]=num[1];
        dpf[1]=sumf[1];
        for(int i=2;i<=N ;i++ ){
            sumf[i]=sumf[i-1]+num[i]>num[i]?sumf[i-1]+num[i]:num[i];
            dpf[i]=max(sumf[i],dpf[i-1]);
        }
        sumr[N]=num[N];
        dpr[N]=sumr[N];
        for(int i=N-1;i>=1 ;i-- ){
            sumr[i]=sumr[i+1]+num[i]>num[i]?sumr[i+1]+num[i]:num[i];
            dpr[i]=max(sumr[i],dpr[i+1]);
        }
        int ans=-0x3f3f3f3f;
        for(int i=1;i<=N-1 ;i++ ){
            ans=max(ans,dpf[i]+dpr[i+1]);
        }
        printf("%d
",ans);
    }
}