This article is made by Jason-Cow.

Welcome to reprint.

But please post the writer's address.

连临街表都不用，每个节点的出度为1，数组搞定访问

#include 
     
      #include 
      
       using namespace std;#define N 200010int GI(){  int x=0,c=getchar(),f=0;  while(c<'0'||c>'9'){
    if(c=='-')f=1;c=getchar();}  while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();  return f?-x:x;}int dfn[N],low[N],s[N],in[N],idx,top,ans=1<<30,to[N];void tarjan(int u){    dfn[u]=low[u]=++idx;    s[++top]=u,in[u]=1;    int v=to[u];    if(!dfn[v])tarjan(v),low[u]=min(low[u],low[v]);    else if(in[v])       low[u]=min(low[u],dfn[v]);    if(low[u]==dfn[u]){        int sz=0;        do++sz,in[top]=0;        while(s[top--]!=u);        if(sz>1)ans=min(ans,sz);    }}int main(){    int n=GI(),u,v;    for(v=1;v<=n;v++)to[v]=GI();    for(u=1;u<=n;u++)tarjan(u);    return printf("%d",ans),0;}