MAX Average Problem（斜率优化dp）

MAX Average Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7574    Accepted Submission(s): 1667

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

 

题解：

http://www.docin.com/p-47950655.html这篇论文讲的斜率优化，讲的很清楚；

给定一个长度为n的序列，从其中找连续的长度大于m的子序列使得子序列中的平均值最小。

总结就是：构造下凸折线，维护下凸折线（凸包维护），找与下凸折线相切的斜率（也可以用二分来找）

没用二分：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x) scanf("%lf",&x)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=1e5+;
int sum[MAXN],s[MAXN],a[MAXN];
bool cross(int i,int j,int k){
    if((sum[j]-sum[i])*(k-i)>=(sum[k]-sum[i])*(j-i))return true;
    return false;
}
double flx(int i,int t){
    double temp;
    temp=1.0*(sum[t]-sum[i])/(t-i);
    return temp;
}
int main(){
    int N,k;
    while(~scanf("%d%d",&N,&k)){
        sum[]=;
        for(int i=;i<=N;i++)SI(a[i]),sum[i]=sum[i-]+a[i];
        int top=,low=;
        double ans=;
        for(int i=k;i<=N;i++){
            int j=i-k;
            while(top-low>=&&cross(s[top-],s[top],j))top--;
                s[++top]=j;
            while(top-low>=&&flx(s[low+],i)>=flx(s[low],i))low++;
            ans=max(ans,flx(s[low],i));
        }
        printf("%.2lf\n",ans);
    }
    return ;
}

用了二分：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x) scanf("%lf",&x)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=1e5+;
int sum[MAXN],s[MAXN],a[MAXN];
bool cross(int i,int j,int k){
    if((sum[j]-sum[i])*(k-i)>=(sum[k]-sum[i])*(j-i))return true;
    return false;
}
double flx(int i,int t){
    double temp;
    temp=1.0*(sum[t]-sum[i])/(t-i);
    return temp;
}
int erfen(int l,int r,int i){
    int mid;
    while(l<=r){
        mid=(l+r)>>;
        if(cross(s[mid],s[mid+],i))r=mid-;
        else l=mid+;
    }
    return r+;
}
int main(){
    int N,k;
    while(~scanf("%d%d",&N,&k)){
        sum[]=;
        for(int i=;i<=N;i++)SI(a[i]),sum[i]=sum[i-]+a[i];
        int top=,low=;
        double ans=;
        for(int i=k;i<=N;i++){
            int j=i-k;
            while(top-low>=&&cross(s[top-],s[top],j))top--;
                s[++top]=j;
            ans=max(ans,flx(s[erfen(,top,i)],i));
        }
        printf("%.2lf\n",ans);
    }
    return ;
}