hdu2993坡dp+二进制搜索

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 6
6 4 2 10 3 8 5 9 4 1

 

Sample Output

6.50

參考：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 99999999
typedef 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 99999999
typedef 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){//依据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];//这里一定要注意假设切点是最后一个点须要另判,由于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){//依据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);//依据相邻点与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;
}