【HDOJ6604】Blow up the city（支配树）

题意：给定一个n点m边的DAG，将只有入边的点称为周驿东点

q次询问，每次给定a,b两点，询问删去某个点x和其相连的所有边，能使a,b至少其中之一不能到达任何周驿东点的x的个数

n,q<=1e5，m<=2e5

思路：

 

 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
#define N  210000
#define M  1100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=998244353,inv2=(MOD+1)/2;
      double eps=1e-6;
      ll INF=1e14;

vector<int> dom[N],be[N];
int f[N][20],head[N],vet[N],nxt[N],dep[N],ind[N],
    id[N],semi[N],p[N],pa[N],dfn[N],idom[N],mn[N],tot,cnt;

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

void add(int a,int b)
{
    nxt[++tot]=head[a];
    vet[tot]=b;
    head[a]=tot;
}

void dfs(int u)
{
    dfn[u]=++cnt; id[cnt]=u;
    int e=head[u];
    while(e)
    {
        int v=vet[e];
        if(!dfn[v])
        {
            dfs(v);
            pa[dfn[v]]=dfn[u];
        }
        be[dfn[v]].push_back(dfn[u]);
        e=nxt[e];
    }
}

int get(int x)
{
    if(p[x]!=p[p[x]])
    {
        if(semi[mn[x]]>semi[get(p[x])]) mn[x]=get(p[x]);
        p[x]=p[p[x]];
    }
    return mn[x];
}

void LT()
{
    per(i,cnt,2)
    {
        for(int j:be[i]) semi[i]=min(semi[i],semi[get(j)]);
        dom[semi[i]].push_back(i);
        int x=p[i]=pa[i];
        for(int y:dom[x]) idom[y]=(semi[get(y)]<x?get(y):x);
        dom[x].clear();
    }
    rep(i,2,cnt)
    {
        if(idom[i]!=semi[i]) idom[i]=idom[idom[i]];
        dom[id[idom[i]]].push_back(id[i]);
    }
}

void getdep(int u,int fa)
{
    rep(i,1,19) f[u][i]=f[f[u][i-1]][i-1];
    int e=head[u];
    while(e)
    {
        int v=vet[e];
        if(v!=fa)
        {
            f[v][0]=u;
            dep[v]=dep[u]+1;
            getdep(v,u);
        }
        e=nxt[e];
    }
}

int lca(int x,int y)
{
    if(dep[x]<dep[y]) swap(x,y);
    int d=dep[x]-dep[y];
    rep(i,0,19)
     if((d>>i)&1) x=f[x][i];
    per(i,19,0)
     if(f[x][i]!=f[y][i])
     {
         x=f[x][i];
         y=f[y][i];
     }
    if(x==y) return x;
    return f[x][0];
}

int main()
{
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int cas=read();
    while(cas--)
    {
        int n=read(),m=read();
        tot=0;
        rep(i,1,n+1) head[i]=ind[i]=dfn[i]=id[i]=idom[i]=0;
        rep(i,1,m)
        {
            int x=read(),y=read();
            add(y,x);
            ind[x]++;
        }
        int root=n+1;
        rep(i,1,n)
         if(!ind[i]) add(root,i);
        rep(i,1,root)
        {
            dfn[i]=0;
            dom[i].clear();
            be[i].clear();
            p[i]=mn[i]=semi[i]=i;
        }
        cnt=0;
        dfs(root);
        LT();
        tot=0;
        rep(i,1,root) head[i]=0;
        rep(i,1,root)
         if(id[idom[i]]) add(id[idom[i]],id[i]);
        rep(i,1,root) dep[i]=0;
        getdep(root,0);
        int q=read();
        while(q--)
        {
            int x=read(),y=read();
            int t=lca(x,y);
            printf("%d\n",dep[x]+dep[y]-dep[t]);
        }
    }

    return 0;
}