题意见试题传送门
解题思路：显然是题树剖题。
考虑用线段树维护区间端点颜色与颜色数，这样就可以方便的合并，注意查询的时候对端点的特殊处理即可。
时间效率最高为\( O (m \log^{2} n) \).(BZOJ 上 4072ms)
#include <stdio.h>
#define MN 100005
#define Mn (1<<17)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define mid (l+r>>1)
int lc[Mn<<],rc[Mn<<],sum[Mn<<],mark[Mn<<];
int n,q,col[MN],siz[MN],fa[MN],son[MN],dep[MN],top[MN],pos[MN],head[MN],cnt,dfsn,rev[MN];
int to[MN<<],nxt[MN<<];
inline int in(){
int x=;bool f=; char ch=getchar();
while(ch<''||ch>'') f=ch=='-',ch=getchar();
while(ch>=''&&ch<='') x=(x<<)+(x<<)+ch-'',ch=getchar();
return f?-x:x;
}
inline void swp(int &a,int &b){a^=b^=a^=b;}
inline void ins(int x,int y){to[++cnt]=y,nxt[cnt]=head[x],head[x]=cnt;}
inline void dfs1(int u,int f,int d){
fa[u]=f,dep[u]=d,siz[u]=;
for (register int i=head[u]; i; i=nxt[i])
if (to[i]!=f){
dfs1(to[i],u,d+);siz[u]+=siz[to[i]];
if (siz[to[i]]>siz[son[u]]) son[u]=to[i];
}
}
inline void dfs2(int u,int tp){
top[u]=tp;pos[u]=(++dfsn);rev[dfsn]=u;if (son[u]) dfs2(son[u],tp);
for (register int i=head[u]; i; i=nxt[i])
if (to[i]!=fa[u]&&to[i]!=son[u]) dfs2(to[i],to[i]);
}
inline void pushdown(int k){
mark[ls(k)]=mark[rs(k)]=lc[ls(k)]=rc[ls(k)]=lc[rs(k)]=rc[rs(k)]=mark[k];
sum[ls(k)]=sum[rs(k)]=;mark[k]=;
}
inline void combine(int k){
sum[k]=rc[ls(k)]==lc[rs(k)]?sum[ls(k)]+sum[rs(k)]-:sum[ls(k)]+sum[rs(k)];
lc[k]=lc[ls(k)],rc[k]=rc[rs(k)];
}
inline void build(int k,int l,int r){
mark[k]=;
if (l==r){sum[k]=,lc[k]=rc[k]=col[rev[l]];return;}
build(ls(k),l,mid);build(rs(k),mid+,r);combine(k);
}
inline void A(int l,int r,int a,int b,int k,int col){
if (a<=l&&r<=b){mark[k]=lc[k]=rc[k]=col;sum[k]=;return;}
if (mark[k]) pushdown(k);if (a<=mid) A(l,mid,a,b,ls(k),col);
if (b>mid) A(mid+,r,a,b,rs(k),col);combine(k);
}
inline int Qs(int l,int r,int a,int b,int k){
if (l==a&&r==b) return sum[k];if (mark[k]) pushdown(k);
if (b<=mid) return Qs(l,mid,a,b,ls(k));
if (a>mid) return Qs(mid+,r,a,b,rs(k));
return Qs(l,mid,a,mid,ls(k))+Qs(mid+,r,mid+,b,rs(k))-(rc[ls(k)]==lc[rs(k)]);
}
inline int Qc(int l,int r,int x,int k){
if (l==r) return lc[k];if (mark[k]) pushdown(k);
if (x<=mid) return Qc(l,mid,x,ls(k));return Qc(mid+,r,x,rs(k));
}
inline void update(int x,int y,int cl){
while (top[x]!=top[y]){
if (dep[top[x]]<dep[top[y]]) swp(x,y);
A(,n,pos[top[x]],pos[x],,cl);x=fa[top[x]];
}if (dep[x]>dep[y]) swp(x,y);A(,n,pos[x],pos[y],,cl);
}
inline int query(int x,int y){
register int res=;
while(top[x]!=top[y]){
if (dep[top[x]]<dep[top[y]]) swp(x,y);
res+=Qs(,n,pos[top[x]],pos[x],)-(Qc(,n,pos[top[x]],)==Qc(,n,pos[fa[top[x]]],));
x=fa[top[x]];
}if (dep[x]>dep[y]) swp(x,y);res+=Qs(,n,pos[x],pos[y],);return res;
}
void init(){
n=in(),q=in();for (int i=; i<=n; ++i) col[i]=in();
for (register int i=; i<n; ++i){
register int x=in(),y=in();
ins(x,y); ins(y,x);
}dfs1(,,);dfs2(,);build(,,n);
}
void solve(){
while(q--){
register char ch=getchar();while (ch!='C'&&ch!='Q') ch=getchar();
register int x=in(),y=in();
if (ch=='C') update(x,y,in());
else printf("%d\n",query(x,y));
}
}
int main(){init(); solve(); return ;}