（线性dp 最大连续和）POJ 2479 Maximum sum

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44459		Accepted: 13794

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

和poj2593几乎一样。
https://www.cnblogs.com/Weixu-Liu/p/10512447.html
C++代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = ;
int a[maxn],dpl[maxn],dpr[maxn],m1[maxn],m2[maxn];
int Inf = -0x3f3f3f3f;
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        for(int i = ; i <= n; i++){
            scanf("%d",&a[i]);
        }
        memset(dpl,,sizeof(dpl));
        memset(dpr,,sizeof(dpr));
        m1[] = m2[n+] = Inf;
        for(int i = ; i <= n; i++){
            dpl[i] = max(dpl[i-] + a[i],a[i]);
            if(m1[i-] < dpl[i])
                m1[i] = dpl[i];
            else
                m1[i] = m1[i-];
        }
        for(int i = n; i >= ; i--){
            dpr[i] = max(dpr[i+] + a[i],a[i]);
            if(m2[i+] < dpr[i])
                m2[i] = dpr[i];
            else
                m2[i] = m2[i+];
        }
        int maxsum = Inf;
        int tmp[maxn];
        for(int i = ; i <= n-; i++){
            tmp[i] = m1[i] + m2[i+];
            if(maxsum < tmp[i])
                maxsum = tmp[i];
        }
        printf("%d\n",maxsum);
    }
    return ;
}