洛谷P4689 [Ynoi2016]这是我自己的发明 [莫队]

传送门

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ynoi中比较良心不卡常的题。

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思路

没有换根操作时显然可以变成dfs序莫队随便搞。

换根操作时一个子树可以变成两段区间的并集，也随便搞搞就好了。

这题完全不卡常，随便过。

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代码

#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
	using namespace std;
	#define pii pair<int,int>
	#define fir first
	#define sec second
	#define MP make_pair
	#define rep(i,x,y) for (int i=(x);i<=(y);i++)
	#define drep(i,x,y) for (int i=(x);i>=(y);i--)
	#define go(x) for (int i=head[x];i;i=edge[i].nxt)
	#define templ template<typename T>
	#define sz 505050
	typedef long long ll;
	typedef double db;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
	templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
	templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
	templ inline void read(T& t)
	{
		t=0;char f=0,ch=getchar();double d=0.1;
		while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
		while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
		if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
		t=(f?-t:t);
	}
	template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
	char sr[1<<21],z[20];int C=-1,Z=0;
    inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
    inline void print(register int x)
    {
    	if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    	while(z[++Z]=x%10+48,x/=10);
    	while(sr[++C]=z[Z],--Z);sr[++C]='\n';
    }
	void file()
	{
		#ifndef ONLINE_JUDGE
		freopen("a.in","r",stdin);
		#endif
	}
	inline void chktime()
	{
		#ifndef ONLINE_JUDGE
		cout<<(clock()-t)/1000.0<<'\n';
		#endif
	}
	#ifdef mod
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
	ll inv(ll x){return ksm(x,mod-2);}
	#else
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
	#endif
//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

int n,m;
int w[sz],ww[sz],a[sz];

struct E{int t,nxt;}edge[sz<<1];
int head[sz],ecnt;
void make_edge(int f,int t)
{
	edge[++ecnt]=(E){t,head[f]};
	head[f]=ecnt;
	edge[++ecnt]=(E){f,head[t]};
	head[t]=ecnt;
}

int dep[sz],dfn[sz],low[sz],fa[sz][25],T;
#define v edge[i].t
void dfs(int x,int f)
{
	dfn[x]=++T;a[T]=w[x];fa[x][0]=f;dep[x]=dep[f]+1;
	rep(i,1,20) fa[x][i]=fa[fa[x][i-1]][i-1];
	go(x) if (v!=f) dfs(v,x);
	low[x]=T;
}
#undef v
int jump(int x,int to)
{
	drep(i,20,0)
		if (fa[x][i]&&dep[fa[x][i]]>dep[to])
			x=fa[x][i];
	return x;
}

struct hhh
{
	int l1,r1,l2,r2,id;
	hhh(int ll1=0,int rr1=0,int ll2=0,int rr2=0,int idd=0){l1=ll1,r1=rr1,l2=ll2,r2=rr2,id=idd;}
}qq[sz<<2];
int pos[sz],blo;
void init(){blo=sqrt(n);rep(i,0,n) pos[i]=i/blo;}
struct hh{int a,b,id,p;}q[sz<<4];
inline bool cmp(const hh &x,const hh &y){return pos[x.a]==pos[y.a]?((pos[x.a]&1)?x.b<y.b:x.b>y.b):pos[x.a]<pos[y.a];}

ll Ans[sz];
int cnta[sz],cntb[sz];
ll ans;

void solve(int m)
{
	int l1,r1,l2,r2,id,M=0;
    rep(i,1,m)
    {
    	l1=qq[i].l1,r1=qq[i].r1,l2=qq[i].l2,r2=qq[i].r2,id=qq[i].id;
    	if (l1>r1||l2>r2||!r1||!r2) continue;
        if (l1-1&&l2-1) q[++M]=(hh){l1-1,l2-1,id,1};
        if (l1-1) q[++M]=(hh){l1-1,r2,id,-1};
        if (l2-1) q[++M]=(hh){r1,l2-1,id,-1};
        q[++M]=(hh){r1,r2,id,1};
	}
    init();
    sort(q+1,q+M+1,cmp);
    int A=0,B=0;
    rep(i,1,M)
    {
        while (A<q[i].a) ++A,++cnta[a[A]],ans+=cntb[a[A]];
        while (A>q[i].a) ans-=cntb[a[A]],--cnta[a[A]],--A;
        while (B<q[i].b) ++B,++cntb[a[B]],ans+=cnta[a[B]];
        while (B>q[i].b) ans-=cnta[a[B]],--cntb[a[B]],--B;
        Ans[q[i].id]+=ans*q[i].p;
    }
}

int main()
{
    file();
    int x,y,z,rt=1;
    read(n,m);
    rep(i,1,n) read(w[i]),ww[i]=w[i];
    sort(ww+1,ww+n+1);unique(ww+1,ww+n+1);
    rep(i,1,n) w[i]=lower_bound(ww+1,ww+n+1,w[i])-ww;
    rep(i,1,n-1) read(x,y),make_edge(x,y);
    dfs(1,0);
    int M=0,c=0;
    rep(i,1,m)
    {
    	read(z);
    	if (z==1) { read(rt); continue; }
    	read(x,y);++c;
    	#define in(x) dfn[x]<=dfn[rt]&&dfn[rt]<=low[x]
    	if (in(y)) swap(x,y);
    	if (rt==x)
    	{
    		if (rt==y) qq[++M]=hhh(1,n,1,n,c);
    		else if (in(y)) z=jump(rt,y),qq[++M]=hhh(1,n,1,dfn[z]-1,c),qq[++M]=hhh(1,n,low[z]+1,n,c);
    		else qq[++M]=hhh(1,n,dfn[y],low[y],c);
		}
		else if (in(x))
		{
			z=jump(rt,x);int zz;
			if (rt==y) qq[++M]=hhh(1,n,1,dfn[z]-1,c),qq[++M]=hhh(1,n,low[z]+1,n,c);
			else if (in(y))
				zz=jump(rt,y),
				qq[++M]=hhh(1,dfn[z]-1,1,dfn[zz]-1,c),qq[++M]=hhh(1,dfn[z]-1,low[zz]+1,n,c),
				qq[++M]=hhh(low[z]+1,n,1,dfn[zz]-1,c),qq[++M]=hhh(low[z]+1,n,low[zz]+1,n,c);
			else qq[++M]=hhh(dfn[y],low[y],1,dfn[z]-1,c),qq[++M]=hhh(dfn[y],low[y],low[z]+1,n,c);
		}
		else qq[++M]=hhh(dfn[x],low[x],dfn[y],low[y],c);
		#undef in
	}
	solve(M);
	rep(i,1,c) printf("%lld\n",Ans[i]);
	return 0;
}