【LOJ #6101】「2017 山东二轮集训 Day1」第二题（笛卡尔树 DP）

传送门

开始并不会
翻了下代码发现自己脑抽

每次对于最小值分治两边做即可
也就是建出笛卡尔树然后 $dfs$的事
注意处理时每次次数是上取整

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline ll readll(){
    char ch=gc();
    ll res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int N=100005;
int a[N],fa[N],q[N],top,siz[N],n,k;
ll ans,s[N];
vector<int> e[N];
void dfs(int u){
	siz[u]=1;
	for(int &v:e[u])dfs(v),siz[u]+=siz[v],s[u]+=s[v];
	s[u]+=1ll*siz[u]*(a[u]-a[fa[u]]);
	if(s[u]>0){
		ll x=(s[u]-1)/k+1;
		ans+=x,s[u]-=x*k;
	}
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),k=read();
	for(int i=1;i<=n;i++)a[i]=read();
	for(int i=1;i<=n;i++){
		int last=0;
		while(top&&a[q[top]]>a[i])last=q[top--];
		fa[i]=q[top],q[++top]=i;
		if(last)fa[last]=i;
	}
	int rt=0;
	for(int i=1;i<=n;i++){
		if(fa[i])e[fa[i]].pb(i);
		else rt=i;
	}
	dfs(rt);cout<<ans<<'\n';
	return 0;
}