[bzoj4551] [Tjoi2016&Heoi2016]树

 

这道题可以用树链剖分加线段树维护.

但是考虑到这道题的特殊性质,我们可以将操作离线反过来,用并查集维护.

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<string>
#include<iomanip>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define FILE "dealing"
#define up(i,j,n) for(int i=j;i<=n;++i)
#define db double
#define ull unsigned long long
#define eps 1e-10
#define pii pair<int,int>
int read(){
	int x=0,f=1,ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return f*x;
}
const int maxn=200200,mod=(int)(1e9+7+0.1),limit=(int)(1e6+1);
int n,m;
struct node{
	int y,next;
}e[maxn];
int len=0,linkk[maxn];
int fa[maxn],f[maxn];
void insert(int x,int y){
	e[++len].y=y;
	e[len].next=linkk[x];
	linkk[x]=len;
}
int getfa(int x){
	return x==fa[x]?x:fa[x]=getfa(fa[x]);
}
void dfs(int x){
	for(int i=linkk[x];i;i=e[i].next){
		if(e[i].y==f[x])continue;
		f[e[i].y]=x;
		dfs(e[i].y);
	}
}
char ch[maxn];
int c[maxn],ans[maxn],Ci[maxn];
int main(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
	n=read(),m=read();
	up(i,2,n){
		int x=read(),y=read();
		insert(x,y);insert(y,x);
	}
	dfs(1);
	up(i,1,n)fa[i]=f[i];
	fa[1]=f[1]=1;Ci[1]=1;
	up(i,1,m){
		scanf(" %c",&ch[i]);
		c[i]=read();
		if(ch[i]=='C')
			fa[c[i]]=c[i],Ci[c[i]]++;
	}
	for(int i=m;i>=1;i--){
		if(ch[i]=='C'&&!--Ci[c[i]])fa[c[i]]=f[c[i]];
		else ans[i]=getfa(c[i]);
	}
	up(i,1,m)if(ch[i]=='Q')printf("%d\n",ans[i]);
	return 0;
}