给你一张无向图，保证每个点的出边只有一条
求出从每个点出发，直到访问到已经访问过的点时经过的点数
tarjan线缩点，环会被缩成一个强连通分量，不难发现环上的所有点的答案都是环长
最后来一遍dfs，求出所有点到环的距离
第一次dfs没写return，WA的莫名其妙
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int n,x; vei g[maxn],scc[maxn]; int dfn[maxn],low[maxn],timer=0,cnt=0; int c[maxn]; bool ins[maxn]; stack<int> s; int ans[maxn]; void tarjan(int x){ dfn[x]=low[x]=++timer; s.push(x);ins[x]=1; fo(i,0,g[x].size()){ int y=g[x][i]; if(!dfn[y]){ tarjan(y); low[x]=min(low[x],low[y]); } else if(ins[y]) low[x]=min(low[x],dfn[y]); } if(dfn[x]==low[x]){ cnt++; int y; do{ y=s.top(); s.pop(); ins[y]=0; c[y]=cnt,scc[cnt].pub(y); }while(x!=y); } } bool v[maxn]; int dfs(int x){ if(v[x]) return ans[x]; v[x]=1; ans[x]=1; if(g[x][0]==x) return 1; else if(scc[c[x]].size()>1){ ans[x]=scc[c[x]].size(); return ans[x]; } return ans[x]+=dfs(g[x][0]); } void cal(){ re(i,1,n) if(!v[i]) dfs(i); } signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>n; re(i,1,n) cin>>x,g[i].pub(x); re(i,1,n) if(!dfn[i]) tarjan(i); cal(); re(i,1,n) cout<<ans[i]<<endl; return 0; }