Luogu P3572 [POI2014] PTA-Little Bird problem solution (monotonic queue optimization DP)

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We set $f\left[i\right]$ means that the bird is from$1$ tree jumps toiiThe minimum cost of$i$ trees.

Then, the naive $O(n^2)\; The$transformation and equation is$f[i]=f[j]+\left[d\right[j]\le d[i\left]\right]$， 其中$\left[d\right[j]\le d[i\left]\right]$ If the conditions in brackets are met, it is$1$ , otherwise$0$

We can find that for $x<Y$

• 如果 $f[x]<f\left[y\right]$ , then$x\; is$ definitely useless. Because no matter where it is transferred to$y\; is$ no better than$x\; is$ bad, and$x$ will be eliminated first
• Similarly, if $f\left[x\right]==f\left[y\right]$并且$d[x]\le d\left[y\right]$ , then$x$ must be useless

Then, we can use monotonic queue maintenance, the total time complexity is $O\left(n\right)$

#include<cstdio>
#include<iostream>
#include<deque>
#include<cstring>
using namespace std;
const int Maxn=1000000+10;
int a[Maxn],f[Maxn];
int n,m,k;
deque <int> q;
inline int read()
{
    
    
	int s=0,w=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
    
    if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*w;
}
void push(int pos)
{
    
    
	while(q.size())
	{
    
    
		int x=q.back();
		if(f[x]<f[pos] || f[x]==f[pos] && a[x]>a[pos])break;
		q.pop_back();
	}
	q.push_back(pos);
}
void pop(int i)
{
    
    
	while(q.size())
	{
    
    
		if(q.front()>=i-k)break;
		q.pop_front();
	}
}
int main()
{
    
    
//	freopen("in.txt","r",stdin);
	n=read();
	for(int i=1;i<=n;++i)
	a[i]=read();
	m=read();
	while(m--)
	{
    
    
		k=read();
		q.push_back(1);
		for(int i=2;i<=n;++i)
		{
    
    
			pop(i);
			int j=q.front();
			f[i]=f[j];
			if(a[j]<=a[i])++f[i];
			push(i);
		}
		printf("%d\n",f[n]);
		
		q.clear();
		memset(f,0,sizeof(f));
	}
	return 0;
}