题目思路:

线段树+树链剖分，处理好线段树合并的部分，注意在进行树链求和操作时，对于多部分链的端点需要进行合并判断，如果颜色相同,合并后的答案等于ans-1,反则等于ans

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn=1e5+5;
ll head[maxn],cnt,top[maxn],son[maxn],deep[maxn],pre[maxn];
ll color[maxn],sizx[maxn],dfn[maxn],cnx,w[maxn];
struct node
{
    ll to,nex;
} edge[maxn<<3];
void add(ll u,ll v)
{
    edge[cnt].to=v;
    edge[cnt].nex=head[u];
    head[u]=cnt++;
}
void dfs1(ll u,ll fa)
{
    pre[u]=fa;
    deep[u]=deep[fa]+1;
    ll maxson=-1;
    sizx[u]=1;
    for(ll i=head[u]; ~i; i=edge[i].nex)
    {
        ll v=edge[i].to;
        if(v==fa)
            continue;
        dfs1(v,u);
        sizx[u]+=sizx[v];
        if(maxson<sizx[v])
        {
            maxson=sizx[v];
            son[u]=v;
        }
    }
}
void dfs2(ll u,ll t)
{
    top[u]=t;
    dfn[u]=++cnx;
    w[cnx]=color[u];
    if(!son[u])
        return ;
    dfs2(son[u],t);
    for(ll i=head[u]; ~i; i=edge[i].nex)
    {
        ll v=edge[i].to;
        if(v==pre[u]||v==son[u])
        {
            continue;
        }
        dfs2(v,v);
    }
}
struct vain
{
    ll l,r;
    ll sum,lazy,lc,rc;
} tr[maxn<<2];
void pushup(ll k)
{
    ll sum=tr[k<<1].sum+tr[k<<1|1].sum;
    if(tr[k<<1].rc==tr[k<<1|1].lc)
        tr[k].sum=sum-1;
    else
        tr[k].sum=sum;
    tr[k].lc=tr[k<<1].lc;
    tr[k].rc=tr[k<<1|1].rc;
}
void pushdown(ll k)
{
    if(tr[k].lazy)
    {
        tr[k<<1].lc=tr[k<<1].rc=tr[k].lazy;
        tr[k<<1|1].lc=tr[k<<1|1].rc=tr[k].lazy;
        tr[k<<1].lazy=tr[k<<1|1].lazy=tr[k].lazy;
        tr[k<<1].sum=tr[k<<1|1].sum=1;
        tr[k].lazy=0;
    }
}
void build(ll k,ll l,ll r)
{
    tr[k].l=l,tr[k].r=r;
    if(l==r)
    {
        tr[k].lc=tr[k].rc=w[l];
        tr[k].sum=1;
        return ;
    }
    ll mid=l+r>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    pushup(k);
}
void modify(ll k,ll l,ll r,ll c)
{
    if(tr[k].l>=l&&tr[k].r<=r)
    {
        tr[k].lazy=c;
        tr[k].lc=tr[k].rc=c;
        tr[k].sum=1;
        return ;
    }
    pushdown(k);
    ll mid=tr[k].l+tr[k].r>>1;
    if(mid>=r)
        modify(k<<1,l,r,c);
    else if(mid<l)
        modify(k<<1|1,l,r,c);
    else
        modify(k<<1,l,mid,c),modify(k<<1|1,mid+1,r,c);
    pushup(k);
}
ll qid(ll k,ll x)
{
    if(tr[k].l==tr[k].r)
        return tr[k].lc;
    pushdown(k);
    ll mid=tr[k].l+tr[k].r>>1;
    if(mid>=x)
        return qid(k<<1,x);
    else
        return qid(k<<1|1,x);
    pushup(k);
}
ll ask(ll k,ll l,ll r)
{
    if(tr[k].l>=l&&tr[k].r<=r)
        return tr[k].sum;
    pushdown(k);
    ll mid=tr[k].l+tr[k].r>>1;
    if(mid>=r)
        return ask(k<<1,l,r);
    else if(mid<l)
        return ask(k<<1|1,l,r);
    else
        return (ask(k<<1,l,mid)+ask(k<<1|1,mid+1,r))-((tr[k<<1].rc==tr[k<<1|1].lc)?1:0);
}
void opt1(ll x,ll y,ll c)
{

    while(top[x]!=top[y])
    {
        if(deep[top[x]]<deep[top[y]])
            swap(x,y);
        modify(1,dfn[top[x]],dfn[x],c);
        x=pre[top[x]];
    }
    if(deep[x]>deep[y])
        swap(x,y);
    modify(1,dfn[x],dfn[y],c);
}
ll opt2(ll x,ll y)
{
    ll res=0;
    while(top[x]!=top[y])
    {
        if(deep[top[x]]<deep[top[y]])
            swap(x,y);
        res+=ask(1,dfn[top[x]],dfn[x]);
        if(qid(1,dfn[top[x]])==qid(1,dfn[pre[top[x]]]))
            res--;
        x=pre[top[x]];
    }
    if(deep[x]>deep[y])
        swap(x,y);
    res+=ask(1,dfn[x],dfn[y]);
    return res;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    memset(head,-1,sizeof(head));
    ll n,q;
    cin>>n>>q;
    for(ll i=1; i<=n; i++)
    {
        cin>>color[i];
    }
    ll u,v;
    for(ll i=1; i<n; i++)
    {
        cin>>u>>v,add(u,v),add(v,u);
    }
    dfs1(1,0);
    dfs2(1,1);
    build(1,1,n);
    while(q--)
    {
        char s;
        ll x,y,z;
        cin>>s;
        if(s=='Q')
        {
            cin>>x>>y;
            cout<<opt2(x,y)<<endl;
        }
        else
        {
            cin>>x>>y>>z;
            opt1(x,y,z);
        }
    }
}