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~

A  Maximum sum

Accept:40    Submit:103Time 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;}