Poj 2018 Best Cow Fences(分数规划+DP&&斜率优化)

Best Cow Fences

Time Limit: 1000MS Memory Limit: 30000K

Description

Farmer John’s farm consists of a long row of N (1 <= N <= 100,000)fields. Each field contains a certain number of cows, 1 <= ncows <= 2000.

FJ wants to build a fence around a contiguous group of these fields in order to maximize the average number of cows per field within that block. The block must contain at least F (1 <= F <= N) fields, where F given as input.

Calculate the fence placement that maximizes the average, given the constraint.

Input

* Line 1: Two space-separated integers, N and F.

* Lines 2..N+1: Each line contains a single integer, the number of cows in a field. Line 2 gives the number of cows in field 1,line 3 gives the number in field 2, and so on.

Output

* Line 1: A single integer that is 1000 times the maximal average.Do not perform rounding, just print the integer that is 1000*ncows/nfields.

Sample Input

10 6

6

4

2

10

3

8

5

9

4

1

Sample Output

6500

Source

USACO 2003 March Green

/*
01分数规划思想+DP做法.
求长度不小于k连续一段的最大平均值.
ans=(sum[j]-sum[i-1])/(j-i+1) [j-i+1>=k]
这样的话我们二分一个ans,然后让每个数都减去ans.
最后DP检验最大的连续和是否大于0即可.
复杂度O(nlogn).
*/
#include<iostream>
#include<cstdio>
#define MAXN 100001
#define eps 1e-6
using namespace std;
double ans,s[MAXN],sum[MAXN],max1,min1=1e9;
int n,k;
int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
return x*f;
}
bool check(double x)
{
double tot,m=0;
for(int i=1;i<=n;i++) sum[i]=sum[i-1]+s[i]-x;
for(int i=k;i<=n;i++)
{
tot=sum[i]-m;
if(tot>=0) return true;
m=min(m,sum[i-k+1]);
}
return false;
}
void erfen(double l,double r)
{
double mid;
while(l+eps<=r)
{
mid=(l+r)/2;
if(check(mid)) ans=mid,l=mid;
else r=mid;
}
int x=r*1000;
printf("%d\n",x);
return ;
}
int main()
{
int x;
while(~scanf("%d %d",&n,&k))
{
max1=0,min1=1e9;
for(int i=1;i<=n;i++)
{
s[i]=read();
sum[i]=sum[i-1]+s[i];
max1=max(max1,s[i]),min1=min(min1,s[i]);
}
erfen(min1,max1);
}
return 0;
}

/*
斜率优化题.
通过此题大概知道了斜率优化是什么.
因为ans=(sum[j]-sum[i-1])/(j-i+1) [j-i+1>=k].
所以我们可以将看做二维平面中的两个点(i-1,sum[i])和(j,sum[j]).
ans即为两点之间的斜率.
我们要让ans最大化.
n^2的暴力可以看做是有一个点t,和点集Gt{x,0<=x<=t-k},
扫描G中的每个点,使斜率最大化.
但是有些点对于决策是没有影响的.
现在有一个定理:
存在三个有序点i,j,k，如果满足k(i,j)>k(k,j)
即j点是直线ik上方的点,那它不会对决策产生影响.
证明的话见2004周源国家集训队论文,讲的很清楚.
这样的话我们维护一个下凸折线就好了.
我们每次都加入位置为i-k的点,维护下凸性.
如果某一点j(0<=j<=i-k)与点i的斜率最大,
则这个点显然一定是下凸折线的切点.
则它之前的点连线斜率较小也不会对决策产生影响
删掉就可以了.
这样的话由于每个点只入队出队一次
所以复杂度是O(n)的.
参考资料 2004周源国家集训队论文.
*/
#include<iostream>
#include<cstdio>
#define MAXN 100001
using namespace std;
double ans,sum[MAXN],max1,min1=1e9;
int n,k;
struct data{double x,y;}q[MAXN];
int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
return x*f;
}
double check(data a,data b,data c)
{
return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}
double Get(data a,data b)
{
return (b.y-a.y)/(b.x-a.x);
}
void slove()
{
data x,now;ans=0;
int head=0,tail=0;
for(int i=k;i<=n;i++)
{
x.x=i-k,x.y=sum[i-k];
while(tail-head>=2&&check(q[tail-2],q[tail-1],x)<0) tail--;//维护下凸性.
q[tail++]=x;
now.x=i,now.y=sum[i];
double k=Get(q[head],now);
while(tail-head>0&&Get(q[head+1],now)>k)//删除没用的点.
{
k=Get(q[head+1],now);
head++;
}
ans=max(ans,k);
}
ans*=1000;
int a=ans;
printf("%d\n",a);
}
int main()
{
int x;
while(~scanf("%d %d",&n,&k))
{
for(int i=1;i<=n;i++)
x=read(),sum[i]=sum[i-1]+x;
slove();
}
return 0;
}