poj2479：Maximum sum

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.

题解

寻找最大连续子段和，因为求两个所以正反向各一遍，再遍历分界点。

先求到当前位置的最大子段和（包含该位置），再求当前位置及之前的最大子段和（可包含可不包含）

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 50005
#define inf 1<<29
int f[maxn],d[maxn],a[maxn],lf[maxn],rf[maxn];
int ans;
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ans=-inf;
        for(int i=1 ; i<=n ; ++i )
            scanf("%d",&a[i]);
        f[0]=f[n+1]=d[0]=d[n+1]=lf[0]=lf[n+1]=rf[0]=rf[n+1]=-inf;
        for(int i=1 ; i<=n ; ++i)
            f[i]=max(a[i],f[i-1]+a[i]);
        for(int i=1 ; i<=n ; ++i)
            lf[i]=max(lf[i-1],f[i]);
        for(int i=n ; i>=1 ; --i)
            d[i]=max(a[i],d[i+1]+a[i]);
        for(int i=n ; i>=1 ; --i)
            rf[i]=max(rf[i+1],d[i]);
        for(int i=1 ; i<n ; ++i )
            ans=max(ans,lf[i]+rf[i+1]);
        printf("%d\n",ans);
    }
    return 0;
}