[CSP校内集训]attack（DAG支配树）

题意

给一个DAG，多次询问，每次给定\(k\)个点，求1到这些点的必经点的交集大小

思路

支配树裸题，建好DAG的支配树后\(k\)个点LCA的深度即为答案

Code

#include<bits/stdc++.h>
#define N 100005
using namespace std;
int n,m,q;
int rd[N],f[N][18],dep[N];

struct Edge
{
	int next,to;
}edge[N<<1],edge1[N<<1];int head[N],head1[N],cnt,cnt1;
void add_edge(int from,int to) {edge[++cnt].next=head[from]; edge[cnt].to=to; head[from]=cnt;}
void add_edge1(int from,int to) {edge1[++cnt1].next=head1[from]; edge1[cnt1].to=to; head1[from]=cnt1;}

template <class T>
void read(T &x)
{
	char c;int sign=1;
	while((c=getchar())>'9'||c<'0') if(c=='-') sign=-1; x=c-48;
	while((c=getchar())>='0'&&c<='9') x=x*10+c-48; x*=sign;
}
int lca(int x,int y)
{
	if(dep[x]<dep[y]) swap(x,y);
	for(int i=17;i>=0;--i) if(dep[f[x][i]]>=dep[y]) x=f[x][i];
	if(x==y) return x;
	for(int i=17;i>=0;--i) if(f[x][i]!=f[y][i]) x=f[x][i],y=f[y][i];
	return f[x][0];
}
void toposort()
{
	queue<int> q;
	q.push(1);
	while(!q.empty())
	{
		int u=q.front(),now=-1; q.pop();
		for(int i=head1[u];i;i=edge1[i].next)
		{
			int v=edge1[i].to;
			if(now==-1) now=v;
			else now=lca(now,v);
		}
		if(now==-1) dep[u]=1;
		else
		{
			dep[u]=dep[now]+1;
			f[u][0]=now;
			for(int i=1;i<=17;++i) f[u][i]=f[f[u][i-1]][i-1];
		}
		
		for(int i=head[u];i;i=edge[i].next)
		{
			int v=edge[i].to;
			if(--rd[v]==0) q.push(v);
		}
	}
}
int main()
{
	freopen("attack.in","r",stdin);
	freopen("attack.out","w",stdout);
	read(n);read(m);read(q);
	for(int i=1;i<=m;++i)
	{
		int x,y;
		read(x);read(y);
		add_edge(x,y);
		add_edge1(y,x);
		++rd[y];
	}
	toposort();
	while(q--)
	{
		int k; read(k);
		int x,y; read(x);
		for(int i=2;i<=k;++i)
		{
			read(y);
			x=lca(x,y);
		}
		printf("%d\n",dep[x]);
	}
	return 0;
}