这道题目是维护边权信息，于是我就把原树中边变成点维护。由于颜色不超过30种，可以每个点维护一个int整数代表自己子树的颜色集合，pushup时并(就是 | 这个运算符)一下即可。对于每个原树中的点，颜色一定要保持为0。pushdown时，要判断一下，使得原树中的点颜色始终保持为0，不能被修改，不然就会出错。具体实现见代码。
接下来就好办了，操作1就先cut再link，操作2就先提取这段路径再打懒标，操作3就先提取路径再求一下颜色总数即可。
至于操作1的判断祖孙关系，暂时不知道怎么快速判断，于是我只好在原树中暴力判断，结果居然过了。。。事实上，当树变得很高时，我的打法会被卡。
代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=50005;
int n,m,op,u,v,c,cnt,ans,pa[N],rt[N];
int fa[N*2],ch[N*2][2],rev[N*2],tag[N*2],sumv[N*2],siz[N*2],val[N*2],stk[N*2];
int head[N*2],to[N*4],nxt[N*4];
void adde(int u,int v){to[++cnt]=v;nxt[cnt]=head[u];head[u]=cnt;
}
void dfs(int u){int v;for(int i=head[u];i;i=nxt[i]){v=to[i];if(v!=fa[u]){fa[v]=u;dfs(v);}}
}
bool isroot(int u){return ch[fa[u]][0]!=u&&ch[fa[u]][1]!=u;
}
int which(int u){return u==ch[fa[u]][1];
}
void pushup(int u){sumv[u]=1<<val[u];siz[u]=1;if(ch[u][0]){sumv[u]|=sumv[ch[u][0]];siz[u]+=siz[ch[u][0]];}if(ch[u][1]){sumv[u]|=sumv[ch[u][1]];siz[u]+=siz[ch[u][1]];}
}
void reverse(int u){rev[u]^=1;swap(ch[u][0],ch[u][1]);
}
void maintain(int u,int c){tag[u]=c;sumv[u]=1<<c;if(u>n){val[u]=c;}
}
void downtag(int u){if(rev[u]){if(ch[u][0]){reverse(ch[u][0]);}if(ch[u][1]){reverse(ch[u][1]);}rev[u]=0;}if(tag[u]){if(ch[u][0]){maintain(ch[u][0],tag[u]);}if(ch[u][1]){maintain(ch[u][1],tag[u]);}tag[u]=0;}
}
void pushdown(int u){stk[stk[0]=1]=u;for(;!isroot(u);u=fa[u]){stk[++stk[0]]=fa[u];}while(stk[0]){downtag(stk[stk[0]--]);}
}
void rotate(int x){int y=fa[x],z=fa[y],md=which(x);if(!isroot(y)){ch[z][which(y)]=x;}fa[x]=z;ch[y][md]=ch[x][!md];fa[ch[y][md]]=y;ch[x][!md]=y;fa[y]=x;pushup(y);pushup(x);
}
void splay(int u){pushdown(u);while(!isroot(u)){if(!isroot(fa[u])){rotate(which(fa[u])==which(u)?fa[u]:u);}rotate(u);}
}
void access(int u){for(int v=0;u;v=u,u=fa[u]){splay(u);ch[u][1]=v;pushup(u);}
}
void makeroot(int u){access(u);splay(u);reverse(u);
}
void link(int u,int v){makeroot(u);fa[u]=v;
}
void cut(int u,int v){makeroot(u);access(v);splay(v);fa[u]=ch[v][0]=0;pushup(v);
}
/*bool judge(int u,int v){//用这个判断祖孙关系是错的if(u==v){return false;}access(v);splay(v);splay(u);return fa[v]==0;
}*/
bool judge(int u,int v){//暴力判断祖孙关系while(v){if(u==v){return false;}v=pa[v];}return true;
}
bool isconnect(int u,int v){if(u==v){return true;}makeroot(u);access(v);splay(v);return fa[u]!=0;
}
int main(){while(~scanf("%d%d",&n,&m)){memset(fa,0,sizeof(fa));memset(ch,0,sizeof(ch));memset(rev,0,sizeof(rev));memset(tag,0,sizeof(tag));memset(sumv,0,sizeof(sumv));memset(siz,0,sizeof(siz));memset(val,0,sizeof(val));memset(head,0,sizeof(head));rt[0]=cnt=0;for(int i=1;i<=n;i++){scanf("%d",&pa[i]);if(pa[i]){adde(pa[i],i+n);adde(i+n,pa[i]);adde(i,i+n);adde(i+n,i);}else{rt[++rt[0]]=i;}}for(int i=1;i<=n;i++){scanf("%d",&val[i+n]);sumv[i+n]=1<<val[i+n];siz[i]=siz[i+n]=1;}for(int i=1;i<=rt[0];i++){dfs(rt[i]);}for(int i=1;i<=m;i++){scanf("%d%d%d",&op,&u,&v);if(op==1){scanf("%d",&c);if(judge(u,v)){if(pa[u]){cut(pa[u],u+n);cut(u+n,u);}pa[u]=v;link(pa[u],u+n);link(u+n,u);splay(u+n);val[u+n]=c;pushup(u+n);}}else if(op==2){scanf("%d",&c);if(u!=v&&isconnect(u,v)){makeroot(u);access(v);splay(v);maintain(v,c);}}else{if(u==v||!isconnect(u,v)){puts("0 0");}else{makeroot(u);access(v);splay(v);ans=0;for(int j=1;j<=30;j++){if(sumv[v]&(1<<j)){ans++;}}printf("%d %d\n",(siz[v]-1)/2,ans);}}}}return 0;
}