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\(s_i=\sum\limits_ {j=1}^ia_ j\)

显然有\(a_ i=\sum\limits_ {j=1}^id_ j\)

于是大力展开得到

\[s_i=\sum\limits_{j=1}^i(i-j+1)d_j\]

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

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
,he
,ne[M],to[M],f
,s
,id
,pr
,dep
,top
,d
,e
;
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('\n');
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('\n');
}
fwrite(obuf,1,op-obuf,stdout);
return 0;
}