【刷题】BZOJ 4543 [POI2014]Hotel加强版

Description

同OJ3522

数据范围：n<=100000

Solution

dp的设计见【刷题】BZOJ 3522 [Poi2014]Hotel

然后发现dp的第二维与深度有关，于是长链剖分就可以优化成 \(O(n)\) 的了

不会写指针，所以写deque

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define REP(a,b,c) for(register int a=b,a##end=c;a<=a##end;++a)
#define DEP(a,b,c) for(register int a=b,a##end=c;a>=a##end;--a)
const int MAXN=100000+10;
int n,e,to[MAXN<<1],nex[MAXN<<1],beg[MAXN],dep[MAXN],Mxdep[MAXN],hson[MAXN],top[MAXN],id[MAXN],cnt;
ll ans;
std::deque<int> f[MAXN],g[MAXN];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void insert(int x,int y)
{
	to[++e]=y;
	nex[e]=beg[x];
	beg[x]=e;
}
inline void dfs1(int x,int p)
{
	dep[x]=dep[p]+1;Mxdep[x]=dep[x];hson[x]=x;
	for(register int i=beg[x];i;i=nex[i])
		if(to[i]==p)continue;
		else
		{
			dfs1(to[i],x);
			if(Mxdep[to[i]]>Mxdep[x])Mxdep[x]=Mxdep[to[i]],hson[x]=to[i];
		}
}
inline void dfs2(int x,int p,int tp)
{
	top[x]=tp;
	if(hson[x]!=x)dfs2(hson[x],x,tp);
	for(register int i=beg[x];i;i=nex[i])
		if(to[i]==p||to[i]==hson[x])continue;
		else dfs2(to[i],x,to[i]);
}
inline void dfs(int x,int p)
{
	if(hson[x]==x)
	{
		id[x]=++cnt;
		f[id[x]].resize(dep[x]-dep[top[x]]+1);
		g[id[x]].resize(dep[x]-dep[top[x]]+1);
		f[id[x]][0]=1;
		return ;
	}
	dfs(hson[x],x);
	id[x]=id[hson[x]];
	f[id[x]].push_front(1);f[id[x]].pop_back();
	g[id[x]].push_back(0);g[id[x]].pop_front();
	ans+=g[id[x]][0];
	for(register int i=beg[x];i;i=nex[i])
		if(to[i]==p||to[i]==hson[x])continue;
		else
		{
			dfs(to[i],x);
			REP(j,0,Mxdep[to[i]]-dep[to[i]]+1)
			{
				ans+=f[id[x]][j]*g[id[to[i]]][j+1]+(j?f[id[to[i]]][j-1]*g[id[x]][j]:0);
				g[id[x]][j]+=g[id[to[i]]][j+1]+(j?f[id[x]][j]*f[id[to[i]]][j-1]:0);
				if(j)f[id[x]][j]+=f[id[to[i]]][j-1];
			}
			while(!f[id[to[i]]].empty())f[id[to[i]]].pop_back();
			while(!g[id[to[i]]].empty())g[id[to[i]]].pop_back();
		}
}
int main()
{
	read(n);
	REP(i,1,n-1)
	{
		int u,v;read(u);read(v);
		insert(u,v);insert(v,u);
	}
	dfs1(1,0);dfs2(1,0,1);dfs(1,0);
	write(ans,'\n');
	return 0;
}