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;
}
原文地址:https://www.cnblogs.com/Chtholly/p/11730093.html
时间: 2024-11-08 17:35:48