[BZOJ4539][HNOI2016]树(主席树)

思维难度低，代码难度高。

直接上主席树即可。

代码用时：0.5h。抄一次代码，错误率还是比较低的，只有一个地方接口参数写反了。

#include<cstdio>
#include<algorithm>
#include<iostream>
#define rep(i,l,r) for (int i=l; i<=r; i++)
typedef long long ll;
using namespace std;

const int N=100100,M=2000100;
int n,m,tot,dfn,sm[M],ls[M],rs[M],rt
,lf
,rg
,id
;
struct P{ int id,rt,fa; ll l,r; }a
;

template<typename T>inline void rd(T &x){
int t; char ch;
for (t=0; !isdigit(ch=getchar()); t=(ch=='-'));
for (x=ch-'0'; isdigit(ch=getchar()); x=x*10+ch-'0');
if (t) x=-x;
}

void ins(int x,int &y,int L,int R,int k){
y=++tot; sm[y]=sm[x]+1;
if (L==R) return; int mid=(L+R)>>1;
if (k<=mid) rs[y]=rs[x],ins(ls[x],ls[y],L,mid,k);
else ls[y]=ls[x],ins(rs[x],rs[y],mid+1,R,k);
}

int que(int k,int z){
int L=1,R=n,mid,t,x=rt[lf[k]-1],y=rt[rg[k]];
while (L<R){
mid=(L+R)>>1; t=sm[ls[y]]-sm[ls[x]];
if (z<=t) R=mid,x=ls[x],y=ls[y];
else L=mid+1,z-=t,x=rs[x],y=rs[y];
}
return L;
}

int getid(ll x,int ed){
int L=1,R=ed+1,mid;
while (L+1<R){
mid=(L+R)>>1;
if (a[mid].l<=x) L=mid; else R=mid;
}
return L;
}

struct T{
int tot,fst
,pnt[N<<1],len[N<<1],nxt[N<<1];
int fa
,sz
,son
,anc
;
ll d
;
void add(int x,int y,int z)
{ pnt[++tot]=y; len[tot]=z; nxt[tot]=fst[x]; fst[x]=tot; }

void dfs(int x){
int p; sz[x]=1;
for (p=fst[x]; p; p=nxt
){
int y=pnt[p];
if (y!=fa[x]){
fa[y]=x; d[y]=d[x]+len[p];
dfs(y); sz[x]+=sz[y];
if (sz[y]>sz[son[x]]) son[x]=y;
}
}
}

void nbr(int x,int tp){
lf[x]=rg[x]=++dfn; id[dfn]=x; anc[x]=tp; int p;
if (son[x]) nbr(son[x],tp),rg[x]=rg[son[x]];
for (p=fst[x]; p; p=nxt[p]){
int y=pnt[p];
if (y!=fa[x] && y!=son[x]) nbr(y,y),rg[x]=rg[y];
}
}

void div(int x,int tp){
anc[x]=tp; int p;
if (son[x]) div(son[x],tp);
for (p=fst[x]; p; p=nxt[p]){
int y=pnt[p];
if (y!=fa[x] && y!=son[x]) div(y,y);
}
}

int lca(int x,int y){
for (; anc[x]!=anc[y]; x=fa[anc[x]])
if (d[anc[x]]<d[anc[y]]) swap(x,y);
return (d[x]<d[y]) ? x : y;
}

int gettp(int x,int y){
int z;
for (; anc[x]!=anc[y]; x=fa[anc[x]]) z=anc[x];
return (x==y) ? z : son[y];
}

void build(){
rep(i,1,n) ins(1,n,rt[i-1],rt[i],id[i]);
}

ll dist(int x,int y){ return d[x]+d[y]-(d[lca(x,y)]<<1); }
}t1,t2;

int main(){
freopen("bzoj4539.in","r",stdin);
freopen("bzoj4539.out","w",stdout);
rd(n); rd(m); int cas,z; ll x,y;
rd(cas);
rep(i,1,n-1) rd(x),rd(y),t1.add(x,y,1),t1.add(y,x,1);
t1.dfs(1); t1.nbr(1,1);
rep(i,1,n) ins(rt[i-1],rt[i],1,n,id[i]);
a[1].id=1; a[1].rt=1; a[1].l=1; a[1].r=n;
rep(i,1,m){
rd(x); rd(y);
a[i+1].rt=x; a[i+1].id=i+1;
a[i+1].l=a[i].r+1; a[i+1].r=a[i].r+t1.sz[x];
z=getid(y,i); a[i+1].fa=y=que(a[z].rt,y-a[z].l+1);
t2.add(z,i+1,t1.d[y]-t1.d[a[z].rt]+1);
}
t2.dfs(1); t2.div(1,1); int u,v,w; ll ans;
while (cas--){
rd(x); rd(y); u=getid(x,m+1); v=getid(y,m+1); w=t2.lca(u,v);
x=que(a[u].rt,x-a[u].l+1); y=que(a[v].rt,y-a[v].l+1);
if (u==v) printf("%lld\n",t1.dist(x,y));
else{
if (u==w) swap(u,v),swap(x,y);
if (v==w){
v=t2.gettp(u,w);
ans=t1.d[x]-t1.d[a[u].rt]+t2.d[u]-t2.d[v];
x=a[v].fa; ans+=t1.dist(x,y)+1;
}else{
ans=t1.d[x]-t1.d[a[u].rt]+t1.d[y]-t1.d[a[v].rt]+t2.dist(u,v);
u=t2.gettp(u,w); v=t2.gettp(v,w);
x=a[u].fa; y=a[v].fa;
ans-=(t1.d[t1.lca(x,y)]-t1.d[a[w].rt])<<1;
}
printf("%lld\n",ans);
}
}
return 0;
}