HDU 2993 MAX Average Problem dp斜率优化

MAX Average Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5803    Accepted Submission(s): 1433

Problem Description

Consider
a simple sequence which only contains positive integers as a1, a2 ...
an, and a number k. Define ave(i,j) as the average value of the sub
sequence ai ... aj, i<=j. Let’s calculate max(ave(i,j)),
1<=i<=j-k+1<=n.

 

Input

There multiple test cases in the input, each test case contains two lines.
The first line has two integers, N and k (k<=N<=10^5).
The second line has N integers, a1, a2 ... an. All numbers are ranged in [1, 2000].

 

Output

For every test case, output one single line contains a real number, which is mentioned in the description, accurate to 0.01.

 

Sample Input

10 6
6 4 2 10 3 8 5 9 4 1

 

Sample Output

6.50

 

 

这道题不是普通的斜率优化，详细题解参见：http://blog.sina.com.cn/s/blog_ad1f8960010174el.html

由于把题目数据范围从10^4提到了5*10^5，我用自己出的数据对拍，网上的标程基本爆掉。

只有一点要注意，就是不加读入优化绝对TLE。一般O(n)的题都是这样。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long double real;
#define MAXN 510000
#define max(x,y) ((x)>(y)?(x):(y))
typedef long long qword;
int num[MAXN];
int sum[MAXN];
int n;
real f[MAXN];
real ans=;
struct Point
{
        qword x,y;
}pl[MAXN];
int topl=-;
qword xmul(Point p1,Point p2,Point p3)
{
        return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x);
}
real get_k(Point p1,Point p2)
{
        return real(p1.y-p2.y)/(p1.x-p2.x);
}
inline int getInt(int &ret)
{
        char ch=getchar();
        ret=;
        while (ch>''||ch<'')ch=getchar();
        while ()
        {
                ret=ret*+ch-'';
                ch=getchar();
                if (ch>''||ch<'')break;
        }
}

int main()
{
        freopen("input.txt","r",stdin);
        int i,j,k;
        while (~scanf("%d%d",&n,&k))
        {
                Point pp;
                for (i=;i<=n;i++)
                {
            //            scanf("%d",&num[i]);
                        getInt(num[i]);
                        sum[i]=sum[i-]+num[i];
                }
                /*
                for (i=1;i<=n;i++)
                {
                        for (j=i-k+1;j>=1;j--)
                        {
                                ans=max(ans,real(sum[i]-sum[j-1])/(i-j+1));
                        }
                }*/
                ans=;
                int ptr=-;
                int now;
                Point pi;
                int head=;
                topl=-;
                for (i=;i<=n;i++)
                {
                        now=i-k;
                        if (now<)continue;
                        pi.x=i;pi.y=sum[i];
                        pp.x=now;
                        pp.y=sum[now];
                        if (topl<=head)
                        {
                                pl[++topl]=pp;
                        }else
                        {
                                while (topl>=head+&&xmul(pl[topl-],pl[topl],pp)<=)
                                {
                                        topl--;
                                }
                                pl[++topl]=pp;
                        }
                        while (ptr==-||(ptr<topl&&get_k(pl[ptr],pi)<=get_k(pl[ptr+],pi)))
                        {
                                ptr++;
                                head=ptr;
                        }
                        if (ans<get_k(pl[ptr],pi))
                        {
                                --++i;
                        }
                        ans=max(ans,get_k(pl[ptr],pi));
                }
                printf("%.2lf\n",(double)ans);
        }
        return ;
}