hdu2993之斜率dp+二分查找

MAX Average Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5825    Accepted Submission(s): 1446

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 66 4 2 10 3 8 5 9 4 1

 

Sample Output

6.50

参考：kuangbin--hdu2993

直接斜率DP:O(N)

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <cmath>#include <map>#include <iomanip>#define INF 99999999typedef long long LL;using namespace std;const int MAX=100000+10;int n,k;int s[MAX],q[MAX];double dp[MAX],sum[MAX];double GetY(int i,int j){return sum[i]-sum[j];}int GetX(int i,int j){return i-j;}double DP(){int head=0,tail=1;q[head]=0;double ans=0;for(int i=1;i<=n;++i)sum[i]=sum[i-1]+s[i]*1.0;for(int i=k;i<=n;++i){int j=i-k;while(head+1<tail && GetY(j,q[tail-1])*GetX(q[tail-1],q[tail-2])<=GetY(q[tail-1],q[tail-2])*GetX(j,q[tail-1]))--tail;q[tail++]=j;while(head+1<tail && GetY(i,q[head])*GetX(i,q[head+1])<=GetY(i,q[head+1])*GetX(i,q[head]))++head;dp[i]=(sum[i]-sum[q[head]])/(i-q[head]);ans=max(ans,dp[i]);}return ans;}int input(){//加速外挂 char ch=' ';int num=0;while(ch<'0' || ch>'9')ch=getchar();while(ch>='0' && ch<='9')num=num*10+ch-'0',ch=getchar();return num;}int main(){while(~scanf("%d%d",&n,&k)){for(int i=1;i<=n;++i)s[i]=input();printf("%0.2lf\n",DP());}return 0;}斜率DP+二分查找:#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <cmath>#include <map>#include <iomanip>#define INF 99999999typedef long long LL;using namespace std;const int MAX=100000+10;int n,k;int s[MAX],q[MAX];LL sum[MAX];LL GetY(int i,int j){return sum[i]-sum[j];}int GetX(int i,int j){return i-j;}LL check(int mid,int i){return GetY(i,q[mid+1])*GetX(q[mid+1],q[mid])-GetY(q[mid+1],q[mid])*GetX(i,q[mid+1]);}int search(int l,int r,int i){//由于斜率单调递增 /*int top=r;while(l<=r){//根据i与mid的斜率 和 i与mid+1的斜率之差求切点if(l == r && l == top)return q[l];//这里一定要注意如果切点是最后一个点需要另判,因为mid+1不存在会出错 int mid=(l+r)>>1;if(check(mid,i)<0)r=mid-1;else l=mid+1;}*/ while(l<r){//根据i与mid的斜率 和 i与mid+1的斜率之差求切点int mid=(l+r)>>1;if(check(mid,i)<0)r=mid;else l=mid+1;}return q[l];}double DP(){int head=0,tail=1,p;q[head]=0;double ans=0,dp;for(int i=1;i<=n;++i)sum[i]=sum[i-1]+s[i];for(int i=k;i<=n;++i){int j=i-k;while(head+1<tail && GetY(j,q[tail-1])*GetX(q[tail-1],q[tail-2])<=GetY(q[tail-1],q[tail-2])*GetX(j,q[tail-1]))--tail;q[tail++]=j;p=search(head,tail-1,i);//根据相邻点与i点的斜率之差二分查找切点 dp=(sum[i]-sum[p])*1.0/(i-p);if(dp>ans)ans=dp;}return ans;}int input(){//加速外挂 char ch=' ';int num=0;while(ch<'0' || ch>'9')ch=getchar();while(ch>='0' && ch<='9')num=num*10+ch-'0',ch=getchar();return num;}int main(){while(~scanf("%d%d",&n,&k)){for(int i=1;i<=n;++i)s[i]=input();printf("%0.2lf\n",DP());}return 0;}