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Problem solution · Go

Codeforces 1527D — MEX Tree

Codeforces 1527D — MEX Tree: a Go solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Depth-first search
Source
EndlessCheng Codeforces Go
Length
108 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Codeforces 1527D — MEX Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 108 lines of Go from the credited upstream file 1527D.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from EndlessCheng Codeforces Go by Σndless (EndlessCheng) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCodeforces 1527D — MEX Tree · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // https://space.bilibili.com/206214func cf1527D(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush() 	var T, n, v, w int	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &n)		g := make([][]int, n)		for i := 1; i < n; i++ {			Fscan(in, &v, &w)			g[v] = append(g[v], w)			g[w] = append(g[w], v)		} 		ans := make([]int, n+1)		dfn := make([]int, n)		size := make([]int, n+1)		t := 0		var dfs func(int, int) int		dfs = func(v, fa int) (s int) {			t++			dfn[v] = t			for _, w := range g[v] {				if w == fa {					continue				}				sz := dfs(w, v)				if v == 0 {					ans[0] += sz * (sz - 1) / 2 // 不含 0				}				s += sz			}			s++			size[v] = s			return		}		dfs(0, -1)		isAncestor := func(f, v int) bool { return dfn[f] < dfn[v] && dfn[v] < dfn[f]+size[f] } 		p, q, ap := 0, 0, 0		s := 1		for _, w := range g[0] {			sz := size[w]			if w == 1 || isAncestor(w, 1) {				ap = w // ap 是 0 的儿子和 1 的祖先（p 的祖先）				sz -= size[1] // 不含 1			}			ans[1] += sz * s // 含 0 不含 1			s += sz		} 		for i := 1; i < n; {			if q == 0 && isAncestor(p, i) { // i 在 p 下面				p = i				for i++; i < n && isAncestor(i, p); i++ { // i 在 0 到 p 的路径上				}				// 含 0~i-1 不含 i				if i == n || isAncestor(p, i) { // i 还在 p 下面					ans[i] = (n - size[ap]) * (size[p] - size[i])				} else {					if i == ap || isAncestor(ap, i) { // LCA(p,i) 是 ap 的后代						ans[i] = (n - size[ap]) * size[p]						break					}					// i 在其它子树					ans[i] = (n - size[ap] - size[i]) * size[p]				}				continue			}			if isAncestor(p, i) { // i 在 p 下面				p = i				for i++; i < n && (isAncestor(i, p) || isAncestor(i, q)); i++ { // i 在 p 到 q 的路径上				}			} else { // i 在 q 下面				q = i				for i++; i < n && (isAncestor(i, p) || isAncestor(i, q)); i++ { // i 在 p 到 q 的路径上				}			}			// 含 0~i-1 不含 i			if i < n && isAncestor(p, i) { // i 在 p 下面				ans[i] = (size[p] - size[i]) * size[q]			} else if i < n && isAncestor(q, i) { // i 在 q 下面				ans[i] = (size[q] - size[i]) * size[p]			} else { // i 在其它子树				ans[i] = size[p] * size[q]				break			}		} 		for _, v := range ans {			Fprint(out, v, " ")		}		Fprintln(out)	}} //func main() { cf1527D(os.Stdin, os.Stdout) }

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