ACM: 动态规划 poj 2479

                                                                      Maximum sum
Description

Given a set of n integers:A={a1, a2,..., an}, we define a function d(A) as below: Your taskis to calculate d(A).

Input

The input consists ofT(<=30) test cases. The number of test cases (T) isgiven in the first line of the input.
Each test case contains two lines. The first line is an integern(2<=n<=50000). The second linecontains n integers: a1, a2, ..., an. (|ai| <=10000).There is an empty line after each case.

Output

Print exactly one line for eachtest 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

题意: 序列分成两部分, 两部分的连续和再加起来最大的值.

解题思路:
         1. 动态规划. max_left[i]表示左边前i个最大连续和.
                     max_right[i]表示右边前i个最大连续和.
代码:
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 50005

int n;
int a[MAX];
int max_left[MAX];
int max_right[MAX];
int maxsize, temp;

inline int max(int a,int b)
{
    return a > b ?  a : b;
}

int main()
{
    int casenum;
//    freopen("input.txt","r",stdin);
    scanf("%d",&casenum);
    while(casenum--)
    {
        scanf("%d",&n);
        memset(max_left,0,sizeof(max_left));
        memset(max_right,0,sizeof(max_right));
        for(int i = 0; i < n; ++i)
        {
            scanf("%d",&a[i]);
        }

        temp = maxsize = -(1<<29);
        for(int i = 0; i < n; ++i)
        {
            if(temp >= 0)
                temp += a[i];
            else
                temp = a[i];
            maxsize = max(maxsize,temp);
            max_left[i] = maxsize;
        }
        
        temp = maxsize = -(1<<29);
        for(int i = n-1; i >= 0; --i)
        {
            if(temp >= 0)
                temp += a[i];
            else
                temp = a[i];
            maxsize = max(maxsize,temp);
            max_right[i] = maxsize;
        }
        
        int result = -(1<<29);
        for(int i = 0; i < n-1; ++i)
        {
            result = max(result,max_left[i]+max_right[i+1]);
        }
        printf("%d\n",result);
    }
    return 0;
}