树剖模板（洛谷P3384 【模板】树链剖分）（树链剖分，树状数组，树的dfn序）

洛谷题目传送门

仍然是一个板子。

不过蒟蒻去学了一下BIT维护区间修改区间求和，常数果真十分优秀

设数列为(a_i)，差分数组(d_ i=a_ i-a_ {i-1})，前缀和(s_i=sumlimits_ {j=1}^ia_ j)

显然有(a_ i=sumlimits_ {j=1}^id_ j)

于是大力展开得到

[s_i=sumlimits_{j=1}^i(i-j+1)d_j ]

[s_i=(i+1)sumlimits_{i=1}^jd_j-sumlimits_{j=1}^ijd_j ]

用BIT维护(d_i)和(id_i)的前缀和即可。

#include<cstdio>
#define RG register
#define R RG int
#define G          if(++ip==ie)fread (ip=ibuf,1,L,stdin)
#define P(C) *op=C;if(++op==oe)fwrite(op=obuf,1,L,stdout)
typedef long long LL;
const int N=1e5+9,M=N<<1,L=1<<19;
char ibuf[L],*ie=ibuf+L,*ip=ie-1,obuf[L],*oe=obuf+L,*op=obuf;
int YL,n,p,a[N],he[N],ne[M],to[M],f[N],s[N],id[N],pr[N],dep[N],top[N],d[N],e[N];
inline int in(){
    G;while(*ip<'-')G;
    R x=*ip&15;G;
    while(*ip>'-'){(x*=10)+=*ip&15;G;}
    return x;
}
void out(R x){
    if(x>9)out(x/10);
    P(x%10|'0');
}
inline void swap(R&x,R&y){
    R z=x;x=y;y=z;
}
void dfs1(R x,R y){
    s[x]=1;
    dep[x]=dep[f[x]=y]+1;
    for(R i=he[x];i;i=ne[i])
        if(to[i]!=y)dfs1(to[i],x);
    s[y]+=s[x];
    if(s[pr[y]]<s[x])pr[y]=x;
}
void dfs2(R x,R y){
    top[x]=y;
    d[id[x]=++p]=a[x];
    if(!pr[x])return;
    dfs2(pr[x],y);
    for(R i=he[x];i;i=ne[i])
        if(!id[to[i]])dfs2(to[i],to[i]);
}
inline void mdf(R*d,R i,R v){
    for(;i<=n;i+=i&-i)(d[i]+=v)%=YL;
}
inline void upd(R l,R r,R v){
    mdf(d,l,v);mdf(e,l,(LL)v*l%YL);v=YL-v;
    mdf(d,r,v);mdf(e,r,(LL)v*r%YL);
}
inline int qry(R*d,R i){
    R v=0;
    for(;i;i-=i&-i)(v+=d[i])%=YL;
    return v;
}
inline int ask(R l,R r){
    return (((LL)(r+1)*qry(d,r)-qry(e,r)-(LL)(l+1)*qry(d,l)+qry(e,l))%YL+YL)%YL;
}
int main(){
    n=in();R m=in(),rt=in(),i,x,y,z,ans;YL=in();
    for(i=1;i<=n;++i)a[i]=in();
    for(i=1;i<n;++i){
        x=in();y=in();
        ne[++p]=he[x];to[he[x]=p]=y;
        ne[++p]=he[y];to[he[y]=p]=x;
    }
    p=0;dfs1(rt,0);dfs2(rt,rt);
    for(i=n;i;--i){
        d[i]=((d[i]-d[i-1])%YL+YL)%YL;
        e[i]=(LL)i*d[i]%YL;
    }
    for(i=1;i<=n;++i){
        if((x=i+(i&-i))>n)continue;
        (d[x]+=d[i])%=YL;(e[x]+=e[i])%=YL;
    }
    while(m--)
        switch(in()){
            case 1:x=in();y=in();z=in()%YL;
                for(;top[x]!=top[y];x=f[top[x]]){
                    if(dep[top[x]]<dep[top[y]])swap(x,y);
                    upd(id[top[x]],id[x]+1,z);
                }
                if(dep[x]<dep[y])swap(x,y);
                upd(id[y],id[x]+1,z);
                break;
            case 2:x=in();y=in();ans=0;
                for(;top[x]!=top[y];x=f[top[x]]){
                    if(dep[top[x]]<dep[top[y]])swap(x,y);
                    ans+=ask(id[top[x]]-1,id[x]);
                }
                if(dep[x]<dep[y])swap(x,y);
                out((ans+ask(id[y]-1,id[x]))%YL);P('
');
                break;
            case 3:x=in();z=in()%YL;
                upd(id[x],id[x]+s[x],z);
                break;
            case 4:x=in();
                out(ask(id[x]-1,id[x]+s[x]-1));P('
');
        }
    fwrite(obuf,1,op-obuf,stdout);
    return 0;
}