Solution -「BZOJ 3779」重组病毒
Description
Link.
Given is a tree. Every node initially has a color which is different from others'. (called \(col_{x}\))
Def: \(\text{dis}(x,y)\): the number of different colors on the simple path between x and y.
Supporting the following operations:
RELEASE x: For each \(y\) on \(\text{path}(x,rt)\), make \(col_{y}\)=a new color which doesn't exist before.RECENTER x: Make \(x\) become the new root after runningRELEASE x.REQUEST x: Print: for each \(y\) in \(\text{subtree}(x)\), the sum of \(\text{dis}(y,rt)\) divided the number of nodes in \(\text{subtree}(x)\).
Solution
Link Cut Tree.
We can know that \(\text{dis}(x,rt)\) is the number of Fake Edges on \(\text{path}(x,rt)\) pluses one.
If we change a Real Edge \((u,v)\), where \(dep_{u}<dep_{v}\) into a Fake Edge, for each node \(x\) in \(\text{subtree}(v)\), \(\text{dis}(x,rt)\) will be decreased by \(1\).
Vice versa.
In order to support such operation: decrease the subtree by \(1\), we can fix the DFS order of the given tree.
However, we also need to change the root. How can we fix the DFS order of the given tree?
Let's have a discussion. Denote \(x\) for the current operating node, \(rt\) for the current root.
- if \(rt=x\): modify the whole tree directly.
- if \(rt\) isn't in \(\text{subtree}(x)\): modify \(\text{subtree}(x)\).
- if \(rt\) is in \(\text{subtree}(x)\): modify \(\text{subtree}(x)\) and cancel the modfication of \(\text{subtree}(rt)\)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
vector<int> e[100010];
int n,m,indfn[100010],outdfn[100010],sjc,fa[100010][20],dep[100010],rtnow=1;
#define check(x,f) ((indfn[x]<indfn[f])|(indfn[x]>outdfn[f])) // check whether x isn't in subtree(f)
void dfs(int x,int las)
{
dep[x]=dep[las]+1,fa[x][0]=las,indfn[x]=++sjc;
for(int i=1;i^20;++i) fa[x][i]=fa[fa[x][i-1]][i-1];
for(unsigned int i=0;i<e[x].size();++i) if(e[x][i]^las) dfs(e[x][i],x);
outdfn[x]=sjc;
}
int getkth(int x,int k)
{
if(k==0) return x;
else
{
for(int i=0;i^20;++i) if((k>>i)&1) x=fa[x][i];
return x;
}
}
struct LinearTree
{
struct node
{
LL val,tag;
}nodes[400010];
void turn(int x,int l,int r)
{
if(nodes[x].tag)
{
int mid=(l+r)>>1;
nodes[x<<1].val+=(mid-l+1)*nodes[x].tag;
nodes[x<<1|1].val+=(r-mid)*nodes[x].tag;
nodes[x<<1].tag+=nodes[x].tag;
nodes[x<<1|1].tag+=nodes[x].tag;
nodes[x].tag=0;
}
}
void ins(int l,int r,int x,int fr,int ba,int val)
{
if(fr>ba||l>ba||r<fr) return;
if(l>=fr&&r<=ba) nodes[x].val+=(r-l+1)*val,nodes[x].tag+=val;
else
{
int mid=(l+r)>>1;
turn(x,l,r);
ins(l,mid,x<<1,fr,ba,val);
ins(mid+1,r,x<<1|1,fr,ba,val);
nodes[x].val=nodes[x<<1].val+nodes[x<<1|1].val;
}
}
LL find(int l,int r,int x,int fr,int ba)
{
if(fr>ba||l>ba||r<fr) return 0;
if(l>=fr&&r<=ba) return nodes[x].val;
else
{
int mid=(l+r)>>1;
turn(x,l,r);
return find(l,mid,x<<1,fr,ba)+find(mid+1,r,x<<1|1,fr,ba);
}
}
void modify(int x,LL val)
{
if(rtnow==x) ins(1,n,1,1,n,val);
else if(check(rtnow,x)) ins(1,n,1,indfn[x],outdfn[x],val);
else
{
int tmp=getkth(rtnow,dep[rtnow]-dep[x]-1);
ins(1,n,1,1,indfn[tmp]-1,val);
ins(1,n,1,outdfn[tmp]+1,n,val);
}
}
}lrt;
struct LinkCutTree
{
#define wis(x) (nodes[nodes[x].fa].ch[1]==(x))
#define isrt(x) ((nodes[nodes[x].fa].ch[0]^(x))&&(nodes[nodes[x].fa].ch[1]^(x)))
struct node
{
int ch[2],fa;
bool rev;
}nodes[100010];
void turn_down(int x)
{
if(nodes[x].rev)
{
swap(nodes[x].ch[0],nodes[x].ch[1]);
if(nodes[x].ch[0]) nodes[nodes[x].ch[0]].rev^=1;
if(nodes[x].ch[1]) nodes[nodes[x].ch[1]].rev^=1;
nodes[x].rev=0;
}
}
void turn_whole(int x)
{
if(!isrt(x)) turn_whole(nodes[x].fa);
turn_down(x);
}
void rotate(int x)
{
int f=nodes[x].fa,ff=nodes[f].fa,t=wis(x);
nodes[x].fa=ff;
if(!isrt(f)) nodes[ff].ch[wis(f)]=x;
nodes[f].ch[t]=nodes[x].ch[t^1];
nodes[nodes[x].ch[t^1]].fa=f;
nodes[x].ch[t^1]=f;
nodes[f].fa=x;
}
void splay(int x)
{
turn_whole(x);
while(!isrt(x))
{
int f=nodes[x].fa;
if(!isrt(f)) rotate((wis(x)^wis(f))?x:f);
rotate(x);
}
}
int findleft(int x)
{
turn_down(x);
while(nodes[x].ch[0]) x=nodes[x].ch[0],turn_down(x);
return x;
}
void access(int x)
{
for(int y=0;x;y=x,x=nodes[x].fa)
{
splay(x);
if(nodes[x].ch[1]) lrt.modify(findleft(nodes[x].ch[1]),1);
if(y) lrt.modify(findleft(y),-1);
nodes[x].ch[1]=y;
}
}
void makert(int x){access(x),splay(x),nodes[x].rev^=1;}
}lct;
char opt[20];
int opx;
template<typename T>
void read(T &hhh)
{
T x=0,f=1;
char c=getchar();
while(c<'0'||c>'9')
{
if(c=='-') f=-1;
c=getchar();
}
while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+(c^'0'),c=getchar();
if(~f) hhh=x;
else hhh=-x;
}
int main()
{
read(n),read(m);
for(int i=1,x,y;i<n;++i)
{
read(x),read(y);
e[x].emplace_back(y);
e[y].emplace_back(x);
}
dfs(1,0);
for(int i=1;i<=n;++i) lrt.ins(1,n,1,indfn[i],indfn[i],dep[i]),lct.nodes[i].fa=fa[i][0];
while(m--)
{
scanf("%s",opt),read(opx);
if(strcmp(opt,"RELEASE")==0) lct.access(opx);
else if(strcmp(opt,"RECENTER")==0) lct.makert(opx),rtnow=opx;
else
{
if(rtnow==opx) printf("%.10f\n",double(lrt.find(1,n,1,1,n))/n);
else if(check(rtnow,opx)) printf("%.10f\n",double(lrt.find(1,n,1,indfn[opx],outdfn[opx]))/(outdfn[opx]-indfn[opx]+1));
else
{
int tmp=getkth(rtnow,dep[rtnow]-dep[opx]-1);
printf("%.10f\n",double(lrt.find(1,n,1,1,indfn[tmp]-1)+lrt.find(1,n,1,outdfn[tmp]+1,n))/(indfn[tmp]+n-outdfn[tmp]-1));
}
}
}
return 0;
}
