POJ 2479 Maximum sum （双向DP）

Maximum sum

http://poj.org/problem?id=2479

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28485		Accepted: 8713

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.

Source

POJ Contest,Author:Mathematica@ZSU

 

 

双向DP求解：

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=50010;

int n,num[N];
int dpl[N],dpr[N];

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        scanf("%d",&num[0]);
        dpl[0]=num[0];
        for(int i=1;i<n;i++){
            scanf("%d",&num[i]);
            if(dpl[i-1]>=0) //从左向右求最大值
                dpl[i]=dpl[i-1]+num[i];
            else
                dpl[i]=num[i];
        }
        int maxv=num[n-1],ans=-999999;
        dpr[n-1]=num[n-1];
        for(int i=n-1;i>=1;i--){
            if(dpr[i]>=0)    //从右向左求最大值
                dpr[i-1]=dpr[i]+num[i-1];
            else
                dpr[i-1]=num[i-1];
            if(maxv<dpr[i])
                maxv=dpr[i];
            if(ans<maxv+dpl[i-1])
                ans=maxv+dpl[i-1];
        }
        printf("%d\n",ans);
    }
    return 0;
}