2012北邮腾讯创新俱乐部热身赛 A

这场比赛打得还是挺开心的，ak了！哈哈哈~~A题是一道dp，复杂度O（n），挺经典的~

AMaximum sum

Accept:40	Submit:103
Time Limit:1000MS	Memory Limit:65536KB

Description

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

d(A)=sum{a[s1]~a[t1]}+sum{a[s2]~a[t2]}

The rule is 1<=s1<=t1<s2<=t2<=n.

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.

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;

#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif

#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a)        for( int i = (a)-1 ; i >= 0 ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define SZ(a)           ((int)a.size())
#define PP(n,m,a)		puts("---");FF(i,n){FF(j,m)cout << a[i][j] << ' ';puts("");}
#define pb              push_back
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-10;
const double pi = acos(-1.0);
const int maxn = 51111;

int a[maxn];
int dp[maxn];
int sum[maxn];
int b[maxn];
int s2[maxn];
int n,T;

void dpstart()
{
    memset(dp,0,sizeof(dp));
    memset(sum,0,sizeof(sum));
    memset(b,0,sizeof(b));
    memset(s2,0,sizeof(s2));
    dp[1] = a[1];
    sum[1] = a[1];
    for(int i=2;i<=n;i++)
    {
        sum[i] = max(sum[i-1]+a[i],a[i]);
        dp[i] = max(dp[i-1],sum[i]);
    }
    b[n] = a[n];
    s2[n]=a[n];
    for(int i=n-1;i>=1;i--)
    {
        s2[i]=max(s2[i+1]+a[i],a[i]);
        b[i]=max(b[i+1],s2[i]);
    }
    int ans = -inf;
    for(int i=1;i<=n-1;i++)
    {
        ans = max(dp[i]+b[i+1],ans);
    }
    cout<<ans<<endl;
    return ;
}

int main()
{
    cin>>T;
    while(T--)
    {
        cin>>n;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            SS(a[i]);
        }
        dpstart();
    }
    return 0;
}