树链剖分

两次dfs把树剖成一段一段连续的链，整体建一次树，用线段树维护区间信息

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int to[200010],nex[200010],head[100010];
int d[100010],fa[100010],son[100010],s[100010];
int top[100010],id[100010],rk[100010];
ll tree[400010],cov[400010],n;
ll p,w[100010];
void build(int l,int r,int now)
{
    if(l==r)
    {
        tree[now]=w[rk[l]];
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,now<<1);
    build(mid+1,r,now<<1|1);
    tree[now]=(tree[now<<1]+tree[now<<1|1])%p;
}
void change(int l,int r,int now,int x,int y,ll k)
{
    tree[now]=(tree[now]+k*(y-x+1)%p)%p;
    if(x==l&&r==y)
    {
        cov[now]=(cov[now]+k)%p;
        return;
    }
    int mid=(l+r)>>1;
    if(y<=mid)
        change(l,mid,now<<1,x,y,k);
    else if(x>mid)
        change(mid+1,r,now<<1|1,x,y,k);
    else
    {
        change(l,mid,now<<1,x,mid,k);
        change(mid+1,r,now<<1|1,mid+1,y,k);
    }
}
ll query(int l,int r,int now,int x,int y,ll k)
{
    if(x<=l&&r<=y)
    {
        return (tree[now]+k*(r-l+1)%p)%p;
    }
    int mid=(l+r)>>1;
    ll ans=0;
    if(x<=mid)
        ans=(ans+query(l,mid,now<<1,x,y,(k+cov[now])%p)%p)%p;
    if(y>mid)
        ans=(ans+query(mid+1,r,now<<1|1,x,y,(k+cov[now])%p)%p)%p;
    return ans;
}
void dfs1(int now,int deep)
{
    d[now]=deep;
    s[now]=1;
    for(int i=head[now];i>0;i=nex[i])
    {
        if(fa[now]==to[i])
            continue;
        fa[to[i]]=now;
        dfs1(to[i],deep+1);
        s[now]+=s[to[i]];
        if(s[to[i]]>s[son[now]])
            son[now]=to[i];
    }
}
void dfs2(int now,int t)
{
    top[now]=t;
    id[now]=++id[0];
    rk[id[now]]=now;
    if(!son[now])
        return;
    dfs2(son[now],t);
    for(int i=head[now];i>0;i=nex[i])
    {
        if(to[i]==fa[now]||to[i]==son[now])
            continue;
        dfs2(to[i],to[i]);
    }
}
ll lca(int x,int y)
{
    ll ans=0;
    while(top[x]!=top[y])
    {
    if(d[top[x]]>d[top[y]])
    {
        ans=(ans+query(1,n,1,id[top[x]],id[x],0))%p;
        x=fa[top[x]];
    }
    else
    {
        ans=(ans+query(1,n,1,id[top[y]],id[y],0))%p;
        y=fa[top[y]];
    }
    }
    if(d[x]>d[y])
    ans=(ans+query(1,n,1,id[y],id[x],0))%p;
    else
    ans=(ans+query(1,n,1,id[x],id[y],0))%p;
    return ans;
}
void add(int x,int y,ll k)
{
    while(top[x]!=top[y])
    {
    if(d[top[x]]>d[top[y]])
    {
        change(1,n,1,id[top[x]],id[x],k);
        x=fa[top[x]];
    }
    else
    {
        change(1,n,1,id[top[y]],id[y],k);
        y=fa[top[y]];
    }
    }
    if(d[x]>d[y])
    change(1,n,1,id[y],id[x],k);
    else
    change(1,n,1,id[x],id[y],k);
}
/*void show(int l,int r,int now,int k)
{
    if(l==r)
    {
        cout<<rk[l]<<' '<<tree[now]+k<<endl;
        return;
    }
    int mid=(l+r)>>1;
    show(l,mid,now<<1,k+cov[now]);
    show(mid+1,r,now<<1|1,k+cov[now]);
}*/
int main()
{
    //freopen("1.txt","r",stdin);
    int m,r,con=1;
    scanf("%d%d%d%lld",&n,&m,&r,&p);
    for(int i=1;i<=n;++i)
    {
        scanf("%lld",&w[i]);
        w[i]%=p;
    }
    for(int i=1;i<n;++i)
    {
    int a,b;
    scanf("%d%d",&a,&b);
    to[con]=b;
    nex[con]=head[a];
    head[a]=con++;
    to[con]=a;
    nex[con]=head[b];
    head[b]=con++;
    }
    fa[r]=r;
    dfs1(r,1);
    dfs2(r,r);
    build(1,n,1);
    while(m--)
    {
        int op;
        scanf("%d",&op);
        if(op==1)
        {
            int a,b;
            ll c;
            scanf("%d%d%lld",&a,&b,&c);
            c%=p;
            add(a,b,c);
        }
        else if(op==2)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            printf("%lld\n",lca(a,b)%p);
        }
        else if(op==3)
        {
            int a;
            ll k;
            scanf("%d%lld",&a,&k);
            k%=p;
            change(1,n,1,id[a],id[a]+s[a]-1,k);
        }
        else
        {
            int a;
            scanf("%d",&a);
            printf("%lld\n",query(1,n,1,id[a],id[a]+s[a]-1,0)%p);
        }
    }
    return 0;
}