hdu 6604 DAG上的支配树(灭绝树)

Posted 2022-06-02 nervendnig

http://acm.hdu.edu.cn/showproblem.php?pid=6604

很裸的支配树/灭绝树题

一般图的tarjan算法的话,先建立,反向图,然后建立一个超级源点,然后连到几个起点,跑支配树就行

可惜太慢...过不去

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define endl ‘\n‘
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
using namespace std;//head
const int maxn=1e5+100,maxm=2e5+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k,num[maxn],root;
class graphpublic://@按点@
  struct nodeint to,next;e[maxn];
  int head[maxn],nume,ltop[maxn],fa[maxn],deep[maxn],sz[maxn],son[maxn];
  inline void add(int a,int b)e[++nume]=b,head[a];head[a]=nume;
  void init(int n)rep(i,1,n) head[i]=0;nume=0;
  void dfs1(int now=root,int pre=root,int d=0)
    deep[now]=d,fa[now]=pre,sz[now]=1,son[now]=0;
    forn(i,now,head,e)
      int to=e[i].to;
      if(to!=pre) 
        dfs1(to,now,d+1);
        sz[now]+=sz[to];
        if(sz[to]>sz[son[now]]) son[now]=to;
      
    
  
  void dfs2(int now=root,int pre=root,int sp=root)
    ltop[now]=sp;
      if(son[now])  dfs2(son[now],now,sp);
      forn(i,now,head,e)
        int to=e[i].to;
        if(to!=son[now]&&to!=pre) dfs2(to,now,to);
      
  
  void getchain()dfs1();dfs2();
  int lca(int x,int y)
    for(;ltop[x]!=ltop[y];deep[ltop[x]]>deep[ltop[y]]?x=fa[ltop[x]]:y=fa[ltop[y]]);
    return deep[x]<deep[y]?x:y;
  //@基础部分@
  int query(int a,int b)
    int x=lca(a,b);
    int ans=deep[a]+deep[b]-deep[x];
    return ans;
  
g;
class domtreepublic://@dom为最终的支配树,root为根,cnt为每个点的支配点编号和@
  int dfn[maxn],rev[maxn],anc[maxn];
  int semi[maxn],idom[maxn];
  int fa[maxn],mi[maxn],clo;
  struct nodeint to,next;;
  struct graph
    node e[maxn];int head[maxn],nume;
    void init(int n=maxn-5)nume=0;fill_n(head,n+1,0);
    void add(int a,int b)e[++nume]=b,head[a];head[a]=nume;
  inv,nxt,dom;
  void  init(int n=maxn-5)
    clo=0;
    rep(i,1,n)fa[i]=mi[i]=semi[i]=i,rev[i]=dfn[i]=anc[i]=idom[i]=0;
    nxt.init(n),inv.init(n),dom.init(n);
  
  void add(int a,int b)inv.add(b,a),nxt.add(a,b);
  int find(int now)
    if(fa[now]==now) return now;
    int fx=fa[now],y=find(fa[now]);
    if(dfn[semi[mi[fx]]]<dfn[semi[mi[now]]])
      mi[now]=mi[fx];
    return fa[now]=y;
  
  void tdfs(int now)
    dfn[now]=++clo;rev[clo]=now;
    forn(i,now,nxt.head,nxt.e)if(!dfn[nxt.e[i].to])
      anc[nxt.e[i].to]=now,tdfs(nxt.e[i].to);
  
  void maketree(int root,int n=maxn-5)
    tdfs(root);
    per(i,2,n)
      int now=rev[i],tmp=n;
      forn(i,now,inv.head,inv.e)
        int to=inv.e[i].to;if(!dfn[to]) continue;
        if(dfn[to]<dfn[now]) tmp=min(tmp,dfn[to]);
        else find(to),tmp=min(tmp,dfn[semi[mi[to]]]);
      
      semi[now]=rev[tmp];fa[now]=anc[now];
      dom.add(semi[now],now);
      now=rev[i-1];
      forn(i,now,dom.head,dom.e)
        int to=dom.e[i].to;find(to);
        if(semi[mi[to]]==now) idom[to]=now;
        else idom[to]=mi[to];
      
    
    rep(i,2,n)
      int to=rev[i];
      if(idom[to]!=semi[to]) idom[to]=idom[idom[to]];
    
    dom.init(n);
    rep(i,1,n) if(i!=root)g.add(idom[i],i);
  
tree;
int cntin[maxn];
int main() IO;
  cin>>casn;
  while(casn--)
    cin>>n>>m;
    root=n+1;
    tree.init(root);
    g.init(root);
    rep(i,1,n) cntin[i]=0;
    rep(i,1,m)
      int a,b;cin>>a>>b;
      tree.add(b,a);
      cntin[a]++;
    
    rep(i,1,n)if(!cntin[i])tree.add(root,i);
    tree.maketree(root,root);
    g.getchain();
    cin>>m;
    while(m--)
      int a,b;cin>>a>>b;
      cout<<g.query(a,b)<<endl;
    
  

 所以用针对DAG的拓扑排序+倍增做法,就能1000ms以下ac了(一个题套了两个板子..)

#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define fi first
#define endl ‘\n‘
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(ii,x) for(int ii=head[x];ii;ii=e[ii].next)
#pragma GCC optimize("Ofast")
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
const int maxn=3e5+9,maxm=1e9+10,maxp=20;
using namespace std;
int casn,n,m,k,root;
int cntin[maxn];
struct nodeint to,next;;
class graphpublic:
  node e[maxn];int head[maxn],nume;
  void init(int n=maxn-5)nume=0;fill_n(head,n+1,0);
  void add(int a,int b)e[++nume]=b,head[a];head[a]=nume;
inv,nxt,dom;
class domtreepublic://@DAG版本,dom依旧是最终树@
  int deep[maxn],anc[maxn][maxp],que[maxn];
  vector<int>edge;
  void init(int n=maxn-5)
    inv.init(n),nxt.init(n),dom.init(n);
    edge.clear();
  
  void bit(int &x,int h)
    for(int i=0;h>0;++i)
      if(h&1) x=anc[x][i];
      h>>=1;
    
  
  int lca(int a,int b)
    if(deep[a]<deep[b]) swap(a,b);
    bit(a,deep[a]-deep[b]);
    if(a==b) return a;
    per(i,0,maxp-1) if(anc[a][i]!=anc[b][i])
      a=anc[a][i],b=anc[b][i];
    return anc[a][0];
  
  void tpsort(int n)
    int tp=0,ed=0;
    rep(i,1,n) 
      if(!cntin[i]) 
        que[ed++]=i;
        inv.add(0,i);
        nxt.add(i,0);
        edge.push_back(i);
      
    
    while(ed!=tp)
      int now=que[tp++];
      forn(i,now,inv.head,inv.e)
        int to=inv.e[i].to;
        cntin[to]--;
        if(!cntin[to]) que[ed++]=to,edge.push_back(to);
      
    
  
  void maketree()
    for(auto i:edge)
      int fa=-1;
      forn(j,i,nxt.head,nxt.e)
        int to=nxt.e[j].to;
        if(fa==-1) fa=to;
        else fa=lca(fa,to);
      fa=fa==-1?0:fa;
      deep[i]=deep[fa]+1,anc[i][0]=fa;
      rep(j,1,maxp-1) anc[i][j]=anc[anc[i][j-1]][j-1];
      dom.add(fa,i);
    
  
  int query(int a,int b)return deep[a]+deep[b]-deep[lca(a,b)];
tree;

int main() IO;
  cin>>casn;
  while(casn--)
    cin>>n>>m;
    tree.init(n+1);
    while(m--)
      int a,b;cin>>a>>b;
      nxt.add(a,b);
      inv.add(b,a);
      cntin[a]++;
    
    tree.tpsort(n);
    tree.maketree();
    cin>>m;
    while(m--)
      int a,b;cin>>a>>b;
      cout<<tree.query(a,b)<<endl;