几个树链剖分例题

树链剖分把我写吐了.为了研究这个东西花费了我4天的时间.
后面我的代码是对的,然后用来对拍的所谓”题解”是错的,害得我把随机的数据手膜了好几遍.
一开始有写完过样例,提交直接A.后来越写调的时间越长,用来对拍的别人的程序是错的,我自己都快方了.

Codeforces 343D Water Tree

将一棵子树覆盖为1,将一点到根路径上的点覆盖为0,询问一个点是1还是0.
有非常强大的诡异dfs可以A掉这题.
当然树剖就是裸题了.

#include //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  re0 x=0,f=1;rec c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=x*10+c-'0';
  return x*(f?1:-1);
  }
inline void read(rel &x){
  x=0;re0 f=1;rec c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=x*10+c-'0';
  x*=f?1:-1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return pc(48);
  if (x<0) x=-x,pc('-');
  re0 bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  }
inline char fuhao(){
  rec c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
const int yuzu=5e5;
typedef int fuko[yuzu|10];
vector<int> lj[yuzu|10];
int n,m;

namespace shu_lian_pou_fen{
fuko sz,fa,dep,son,dfn,ord,top;int cnt;

void dfs1(int u,int f){
  fa[u]=f,dep[u]=dep[f]+1,sz[u]=1;
  for (int v:lj[u]) if (v^f){
    dfs1(v,u),sz[u]+=sz[v];
    if (sz[v]>sz[son[u]]) son[u]=v;
    }
  }

void dfs2(int u,int _top){
  top[u]=_top,dfn[u]=++cnt,ord[cnt]=u;
  if (son[u]) dfs2(son[u],_top);
  for (int v:lj[u]){
    if (v^fa[u]&&v!=son[u]) dfs2(v,v);
    }
  }

typedef int yuki[yuzu<<2|13];
struct segtree{
#define le rt<<1
#define ri le|1
#define ls le,l,mid
#define rs ri,mid+1,r
yuki val,lazy;
void build(int rt=1,int l=1,int r=n){
  lazy[rt]=-1;
  if (l==r) val[rt]=0;
  else{
    int mid=l+r>>1;
    build(ls),build(rs);
    val[rt]=0;
    }
  }
void push_down(int rt,int l,int r){
  if (~lazy[rt]){
    lazy[le]=lazy[ri]=lazy[rt];
    val[le]=val[ri]=lazy[rt];
    lazy[rt]=-1;  
    }
  }
void update(int ql,int qr,int v,int rt=1,int l=1,int r=n){
  if (ql>r||qrreturn;
  if (ql<=l&&qr>=r) val[rt]=lazy[rt]=v;
  else{  
    int mid=l+r>>1;
    push_down(rt,l,r);
    update(ql,qr,v,ls),update(ql,qr,v,rs);
    }
  }
int query(int u,int rt=1,int l=1,int r=n){
  if (ur) return 1;
  if (l==r) return val[rt];
  int mid=l+r>>1;
  push_down(rt,l,r);
  return query(u,ls)&query(u,rs);
  }
}my_;

void update(int u,int v){
  for (;top[u]!=top[v];u=fa[top[u]]){
    if (dep[top[u]]0);
    }
  if (dep[u]>dep[v]) swap(u,v);
  my_.update(dfn[u],dfn[v],0);
  }

int main(){
  re0 i;
  n=read();
  for (i=1;iint u=read(),v=read();
    lj[u].push_back(v);
    lj[v].push_back(u);
    }
  dfs1(1,0),dfs2(1,1);
  my_.build();
  for (m=read();m--;){
    int k=read(),u=read();
    if (k==1){
      my_.update(dfn[u],dfn[u]+sz[u]-1,1);
      }
    if (k==2){
      update(1,u);
      }
    if (k==3){
      write(my_.query(dfn[u])),pl;
      }
    }
  }
}

int main(){
shu_lian_pou_fen::main();
}

Luogu P4315 月下”毛景树”

单边修改,两点间边覆盖,两点间边加,两点间最大边权.
边权树剖需要稍微改一下,用一个过程preedge事前准备一下.
用vector不是很好存储边的序号,故此用邻接表来做.
线段树需要维护两个标记.

/*
void preedge()中将边所指两点的dep比较一下,把较深那点的点权赋值为该边的边权.
然后询问修改到最后的时候不要改dfn[u]就可以了.
*/
#include //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  re0 x=0,f=1;rec c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=x*10+c-'0';
  return x*(f?1:-1);
  }
inline void read(rel &x){
  x=0;re0 f=1;rec c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=x*10+c-'0';
  x*=f?1:-1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return pc(48);
  if (x<0) x=-x,pc('-');
  re0 bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  }
}using namespace chtholly;
using namespace std;
const int yuzu=1e5,inf=0x3f3f3f3f;
typedef int fuko[yuzu|10];
struct node{int fr,to,next,w;}edge[yuzu<<1|1];
int n=read(),ecnt;

namespace shu_lian_pou_fen{
fuko fa,dep,sz,top,dfn,ord,son,a,head;int cnt;

void addedge(int u,int v,int w){
  edge[++ecnt]={u,v,head[u],w},head[u]=ecnt;
  }

void dfs1(int u,int f){
  fa[u]=f,sz[u]=1,dep[u]=dep[f]+1;
  for (re0 i=head[u];i;i=edge[i].next){
    int v=edge[i].to;
    if (v^f){
      dfs1(v,u),sz[u]+=sz[v];
      if (sz[v]>sz[son[u]]) son[u]=v;
      }
    }
  }

void dfs2(int u,int _top){
  top[u]=_top,dfn[u]=++cnt,ord[cnt]=u;
  if (son[u]) dfs2(son[u],_top);
  for (re0 i=head[u];i;i=edge[i].next){
    int v=edge[i].to;
    if (v^fa[u]&&v!=son[u]) dfs2(v,v);
    }
  }

typedef int yuki[yuzu<<2|13];
struct segtree{
#define le rt<<1
#define ri le|1
#define ls le,l,mid
#define rs ri,mid+1,r
yuki val,cov,lazy;
void build(int rt=1,int l=1,int r=n){
  cov[rt]=-1,lazy[rt]=0;
  if (l==r) val[rt]=a[l];
  else{
    int mid=l+r>>1;
    build(ls),build(rs);
    val[rt]=max(val[le],val[ri]);
    }
  }
void push_down(int rt,int l,int r){
  if (~cov[rt]){
    cov[le]=cov[ri]=cov[rt];
    val[le]=val[ri]=cov[rt];
    lazy[le]=lazy[ri]=0;
    cov[rt]=-1;
    }
  if (lazy[rt]){
    lazy[le]+=lazy[rt],lazy[ri]+=lazy[rt];
    val[le]+=lazy[rt],val[ri]+=lazy[rt];
    lazy[rt]=0;
    }
  }
void cover(int ql,int qr,int v,int rt=1,int l=1,int r=n){
  if (ql>r||qrreturn;
  if (ql<=l&&qr>=r) val[rt]=cov[rt]=v,lazy[rt]=0;
  else{
    int mid=l+r>>1;
    push_down(rt,l,r);
    cover(ql,qr,v,ls),cover(ql,qr,v,rs);
    val[rt]=max(val[le],val[ri]);
    }
  }
void add(int ql,int qr,int v,int rt=1,int l=1,int r=n){
  if (ql>r||qrreturn;
  if (ql<=l&&qr>=r) val[rt]+=v,lazy[rt]+=v;
  else{
    int mid=l+r>>1;
    push_down(rt,l,r);
    add(ql,qr,v,ls),add(ql,qr,v,rs);
    val[rt]=max(val[le],val[ri]);
    }
  }
int query(int ql,int qr,int rt=1,int l=1,int r=n){
  if (ql>r||qrreturn -inf;
  if (ql<=l&&qr>=r) return val[rt];
  int mid=l+r>>1;
  push_down(rt,l,r);
  return max(query(ql,qr,ls),query(ql,qr,rs));
  }
}my_;

void preedge(){
  for (re0 i=1;i<=ecnt;i+=2){
    int &u=edge[i].fr,&v=edge[i].to;
    if (dep[u]>dep[v]) swap(u,v);
    a[dfn[v]]=edge[i].w;
    }
  }

void change(int k,int w){
  int t=dfn[edge[(k<<1)-1].to];
  my_.cover(t,t,w);
  }

void cover(int u,int v,int w){
  for (;top[u]!=top[v];u=fa[top[u]]){
    if (dep[top[u]]if (dep[u]>dep[v]) swap(u,v);
  my_.cover(dfn[u]+1,dfn[v],w);
  }

void add(int u,int v,int w){
  for (;top[u]!=top[v];u=fa[top[u]]){
    if (dep[top[u]]if (dep[u]>dep[v]) swap(u,v);
  my_.add(dfn[u]+1,dfn[v],w);
  }

int q_max(int u,int v){
  int ans=-inf;
  for (;top[u]!=top[v];u=fa[top[u]]){
    if (dep[top[u]]if (dep[u]>dep[v]) swap(u,v);
  return max(ans,my_.query(dfn[u]+1,dfn[v]));
  }

int main(){
  re0 i;
  for (i=1;iint u=read(),v=read(),w=read();
    addedge(u,v,w),addedge(v,u,w);
    }
  dfs1(1,0),dfs2(1,1); 
  preedge();
  my_.build();
  char c[9];
  for (;scanf("%s",c),c[1]!='t';){
    int u=read(),v=read();
    if (c[1]=='a'){//max
      write(q_max(u,v)),pl;
      }
    if (c[1]=='o'){//cover
      cover(u,v,read());
      }
    if (c[1]=='d'){//add
      add(u,v,read());
      }
    if (c[1]=='h'){//change
      change(u,v);
      }
    }
  }
}

int main(){
shu_lian_pou_fen::main();
}

Luogu P3979 遥远的国度

支持区间覆盖,子树最小值,换根.
其它东西很简单,换根是个什么操作呢?

显然我们不可能对于每一次换根都把整棵树重新暴力剖一遍.
我们可以思考一下.
对于询问的点u,我们考虑它与现在的根nowrt的关系.
如果它和现在的根相同,直接返回整棵树.
其余情况求一下这两点的最近公共祖先.
如果答案是u,把u在nowrt方向上的儿子提起来,发现答案就是整棵树去掉nowrt方向上的儿子的子树的其他部分.
剩下还是u的子树不变.

然后分类讨论就做完了.不要忘记树剖自带求lca的功能,不用再来一个倍增数组了.
最后是这个坑爹的inf,0x3f3f3f3f0x3f3f3f3f不对,21474836472147483647也不对,21000000002100000000反而对了.

#include //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  re0 x=0,f=1;rec c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=x*10+c-'0';
  return x*(f?1:-1);
  }
inline void read(rel &x){
  x=0;re0 f=1;rec c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=x*10+c-'0';
  x*=f?1:-1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return pc(48);
  if (x<0) x=-x,pc('-');
  re0 bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  }
inline char fuhao(){
  rec c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
const int yuzu=1e5,inf=2100000000;
typedef int fuko[yuzu|10];
vector<int> lj[yuzu|10];
int n=read(),m=read();

namespace shu_lian_pou_fen{
fuko fa,sz,son,dep,top,dfn,ord,a;
int cnt,nowrt;

void dfs1(int u,int f){
  fa[u]=f,sz[u]=1,dep[u]=dep[f]+1;
  for (re0 i=0;iint v=lj[u][i];
    if (v^f){
      dfs1(v,u),sz[u]+=sz[v];
      if (sz[v]>sz[son[u]]) son[u]=v;
      }
    }
  }

void dfs2(int u,int _top){
  top[u]=_top,dfn[u]=++cnt,ord[cnt]=u;
  if (son[u]) dfs2(son[u],_top);
  for (re0 i=0;iint v=lj[u][i];
    if (!dfn[v]) dfs2(v,v);
    }
  }

void debug(){
  re0 i;
  for (i=1;i<=n;++i) write(fa[i]),p32;pl;
  for (i=1;i<=n;++i) write(dep[i]),p32;pl;
  for (i=1;i<=n;++i) write(son[i]),p32;pl;
  for (i=1;i<=n;++i) write(sz[i]),p32;pl;
  for (i=1;i<=n;++i) write(top[i]),p32;pl;
  for (i=1;i<=n;++i) write(dfn[i]),p32;pl;
  for (i=1;i<=n;++i) write(a[ord[i]]),p32;pl;
  }

typedef int yuki[yuzu<<2|13];
struct segtree{
#define le rt<<1
#define ri le|1
#define ls le,l,mid
#define rs ri,mid+1,r
yuki val,lazy;
void build(int rt=1,int l=1,int r=n){
  lazy[rt]=-1;
  if (l==r) val[rt]=a[ord[l]];
  else{
    int mid=l+r>>1;
    build(ls),build(rs);
    val[rt]=min(val[le],val[ri]);
    }   
  }
void push_down(int rt,int l,int r){
  if (~lazy[rt]){
    lazy[le]=lazy[ri]=lazy[rt];
    val[le]=val[ri]=lazy[rt];
    lazy[rt]=-1;
    }   
  }
void update(int ql,int qr,int v,int rt=1,int l=1,int r=n){
  if (ql>r||qrreturn;
  if (ql<=l&&qr>=r){
    val[rt]=lazy[rt]=v;
    }else{
    int mid=l+r>>1;
    push_down(rt,l,r);
    update(ql,qr,v,ls),update(ql,qr,v,rs);
    val[rt]=min(val[le],val[ri]);
    }
  }
int query(int ql,int qr,int rt=1,int l=1,int r=n){
  if (ql>r||qrreturn inf;
  if (ql<=l&&qr>=r) return val[rt];
  int mid=l+r>>1;
  push_down(rt,l,r);
  return min(query(ql,qr,ls),query(ql,qr,rs));
  }
}my_;

int q_lca(int u,int v){
  for (;top[u]!=top[v];u=fa[top[u]]){
    if (dep[top[u]]return dep[u]//不断跳到所在链的顶端的父亲,最后看看u,v的深度谁大.

void update(int u,int v,int w){
  for (;top[u]!=top[v];u=fa[top[u]]){
    if (dep[top[u]]if (dep[u]>dep[v]) swap(u,v);
  my_.update(dfn[u],dfn[v],w);
  }

int solve(int u){
  if (u==nowrt) return my_.val[1];
  int lca=q_lca(u,nowrt);
  if (lca==u){
    int v;
    for (re0 i=0;iif (v^fa[u]&&dfn[v]<=dfn[nowrt]&&dfn[v]+sz[v]-1>=dfn[nowrt]) break;
      }
    return min(my_.query(1,dfn[v]-1),my_.query(dfn[v]+sz[v],n));
    }
  return my_.query(dfn[u],dfn[u]+sz[u]-1);
  }

int main(){
  re0 i;
  for (i=1;iint u=read(),v=read();
    lj[u].push_back(v);
    lj[v].push_back(u);
    }
  for (i=1;i<=n;++i) a[i]=read();
  nowrt=read();
  dfs1(1,0),dfs2(1,1);
  my_.build();  
  //debug();
  for (;m--;){
    int op=read(),u=read();
    if (op==1){
      nowrt=u;
      }
    if (op==2){
      int v=read();
      update(u,v,read());
      }
    if (op==3){
      write(solve(u)),pl;
      }
    } 
  }
}

int main(){
shu_lian_pou_fen::main();
}

Luogu P2486 染色

这是洛谷原图.

经典题.区间覆盖不难,难的是询问颜色段数量.

/*
线段树每个节点存储该节点最左边是什么颜色lcol,最右边是什么颜色rcol,从左到右颜色段的数量cnt.
pushup的时候判一下左儿子右边的颜色和右儿子左边的颜色是否相同,如果相同cnt[rt]-1.
询问的时候特判一下,如果ql,qr只覆盖了左右两个儿子其中的一个,直接跳去询问那个儿子.
如果同时覆盖了两个儿子,判断一下左儿子右边的颜色和右儿子左边的颜色,相同-1.
那么在询问两点间的时候,处理两链之间的颜色段需要询问一下top[u]的颜色和fa[top[u]]的颜色,如果相同-1.询问颜色是单点查询了,非常简单.
*/
#include //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  re0 x=0,f=1;rec c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=x*10+c-'0';
  return x*(f?1:-1);
  }
inline void read(rel &x){
  x=0;re0 f=1;rec c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=x*10+c-'0';
  x*=f?1:-1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return pc(48);
  if (x<0) x=-x,pc('-');
  re0 bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  }
inline char fuhao(){
  rec c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
const int yuzu=1e5;
typedef int fuko[yuzu|10];
vector<int> lj[yuzu|10];
int n=read(),m=read();

namespace shu_lian_pou_fen{
fuko sz,fa,dep,son,top,dfn,ord,col;int cnt;

void dfs1(int u,int f){
  fa[u]=f,sz[u]=1,dep[u]=dep[f]+1;
  for (re0 i=0;iint v=lj[u][i];
    if (v^f){
      dfs1(v,u),sz[u]+=sz[v];
      if (sz[v]>sz[son[u]]) son[u]=v;
      }
    }
  }

void dfs2(int u,int _top){
  top[u]=_top,dfn[u]=++cnt,ord[cnt]=u;
  if (son[u]) dfs2(son[u],_top);
  for (re0 i=0;iint v=lj[u][i];
    if (v^fa[u]&&v!=son[u]) dfs2(v,v);
    }
  }

typedef int karen[yuzu<<2|13];
struct segtree{
#define le rt<<1
#define ri le|1
#define ls le,l,mid
#define rs ri,mid+1,r
karen lcol,rcol,lazy,cnt;
void pushup(int rt){
  lcol[rt]=lcol[le],rcol[rt]=rcol[ri];
  cnt[rt]=cnt[le]+cnt[ri]-(rcol[le]==lcol[ri]);
  }
void build(int rt=1,int l=1,int r=n){
  lazy[rt]=-1;
  if (l==r){
    lcol[rt]=rcol[rt]=col[ord[l]];
    cnt[rt]=1;
    }else{
    int mid=l+r>>1;
    build(ls),build(rs);
    pushup(rt);
    }
  }
void push_down(int rt,int l,int r){
  if (~lazy[rt]){
    lazy[le]=lazy[ri]=lazy[rt];
    lcol[le]=rcol[le]=lazy[rt];
    lcol[ri]=rcol[ri]=lazy[rt];
    cnt[le]=cnt[ri]=1;
    lazy[rt]=-1;
    }
  }
void update(int ql,int qr,int v,int rt=1,int l=1,int r=n){
  if (ql>r||qrreturn;
  if (ql<=l&&qr>=r){
    lcol[rt]=rcol[rt]=v;
    lazy[rt]=v;
    cnt[rt]=1;
    }else{
    int mid=l+r>>1;
    push_down(rt,l,r);
    update(ql,qr,v,ls),update(ql,qr,v,rs);
    pushup(rt);
    }
  }
int query(int ql,int qr,int rt=1,int l=1,int r=n){
  if (ql<=l&&qr>=r) return cnt[rt];
  int mid=l+r>>1,ans=0;
  push_down(rt,l,r);
  if (ql<=mid) ans+=query(ql,qr,ls);
  if (qr>mid) ans+=query(ql,qr,rs);
  return ans-(ql<=mid&&qr>mid?lcol[ri]==rcol[le]:0);
  }
int get_col(int x,int rt=1,int l=1,int r=n){
  if (l==r) return lcol[rt];
  int mid=l+r>>1;
  push_down(rt,l,r);
  return mid>=x?get_col(x,ls):get_col(x,rs); 
  }
}my_;

int query(int u,int v){
  int ans=0;
  for (;top[u]!=top[v];u=fa[top[u]]){
    if (dep[top[u]]int tmp=my_.get_col(dfn[top[u]])==my_.get_col(dfn[fa[top[u]]]);
    ans+=my_.query(dfn[top[u]],dfn[u])-tmp;
    }
  if (dep[u]>dep[v]) swap(u,v);
  return ans+my_.query(dfn[u],dfn[v]);
  }

void update(int u,int v,int w){
  for (;top[u]!=top[v];u=fa[top[u]]){
    if (dep[top[u]]if (dep[u]>dep[v]) swap(u,v);
  my_.update(dfn[u],dfn[v],w);
  }

int main(){
  re0 i;
  for (i=1;i<=n;++i) col[i]=read();
  for (i=1;iint u=read(),v=read();
    lj[u].push_back(v);
    lj[v].push_back(u);
    }
  dfs1(1,0),dfs2(1,1);
  my_.build();
  for (;m--;){
    char op=fuhao();
    int u=read(),v=read();
    if (op=='Q'){
      write(query(u,v)),pl;
      }
    else{
      update(u,v,read());
      }
    }
  }
}

int main(){
shu_lian_pou_fen::main();
}

比较忙,故此没有在代码中加注释,抱歉了.