基环树森林,然而我比较菜,直接tarjan找环。
发现缩点之后变成了DAG,每一个点往下走一定会走到一个环,缩点之后搜一遍看看会走到哪个环以及那个环的编号是多少,答案就是环的$siz$$ + $要走的路程。
比较垃圾的我忘记了判重边WA了好多发……
时间复杂度$O(n)$。
Code:
#include <cstdio> #include <cstring> using namespace std; const int N = 1e5 + 5; const int inf = 1 << 30; int n, to[N], dfsc = 0, dfn[N], low[N], cur[N]; int scc = 0, siz[N], top = 0, sta[N], bel[N], dis[N]; int tot = 0, head[N]; bool vis[N], isCur[N]; struct Edge { int to, nxt; } e[N]; inline void add(int from, int ver) { e[++tot].to = ver; e[tot].nxt = head[from]; head[from] = tot; } inline void read(int &X) { X = 0; char ch = 0; int op = 1; for(; ch > '9'|| ch < '0'; ch = getchar()) if(ch == '-') op = -1; for(; ch >= '0' && ch <= '9'; ch = getchar()) X = (X << 3) + (X << 1) + ch - 48; X *= op; } inline int min(int x, int y) { return x > y ? y : x; } inline int max(int x, int y) { return x > y ? x : y; } void tarjan(int x) { dfn[x] = low[x] = ++dfsc; sta[++top] = x, vis[x] = 1; int y = to[x]; if(!dfn[y]) { tarjan(y); low[x] = min(low[x], low[y]); } else if(vis[y]) low[x] = min(low[x], dfn[y]); if(low[x] == dfn[x]) { ++scc; for(; sta[top + 1] != x; --top) { bel[sta[top]] = scc; siz[scc]++; vis[sta[top]] = 0; } } } void dfs(int x) { vis[x] = 1; if(isCur[x]) { dis[x] = 0, cur[x] = x; return; } for(int i = head[x]; i; i = e[i].nxt) { int y = e[i].to; if(!vis[y]) dfs(y); dis[x] = dis[y] + 1, cur[x] = cur[y]; } } int main() { // freopen("testdata.in", "r", stdin); // freopen("mine.out", "w", stdout); read(n); for(int i = 1; i <= n; i++) read(to[i]); for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i); /* for(int i = 1; i <= n; i++) printf("%d ", bel[i]); printf(" "); */ for(int i = 1; i <= n; i++) if(to[i] == i) isCur[bel[i]] = 1; for(int i = 1; i <= scc; i++) if(siz[i] >= 2) isCur[i] = 1; for(int i = 1; i <= n; i++) { if(bel[i] == bel[to[i]]) continue; add(bel[i], bel[to[i]]); } for(int i = 1; i <= scc; i++) if(!vis[i]) dfs(i); /* for(int i = 1; i <= scc; i++) printf("%d ", cur[i]); printf(" "); */ for(int i = 1; i <= n; i++) { if(isCur[bel[i]]) printf("%d ", siz[bel[i]]); else printf("%d ", dis[bel[i]] + siz[cur[bel[i]]]); } return 0; }