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Problem solution · Go

Codeforces 1702G2 — Passable Paths (hard version)

Codeforces 1702G2 — Passable Paths (hard version): 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
111 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 1702G2 — Passable Paths (hard version), 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

  • 111 lines of Go from the credited upstream file 1702G2.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 1702G2 — Passable Paths (hard version) · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"math/bits") // https://space.bilibili.com/206214func cf1702G2(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush() 	var n, v, w, dfn, Q, k, aq int	Fscan(in, &n)	g := make([][]int, n+1)	for i := 1; i < n; i++ {		Fscan(in, &v, &w)		v--		w--		g[v] = append(g[v], w)		g[w] = append(g[w], v)	} 	nodes := make([]struct{ l, r int }, n)	const mx = 18	pa := make([][mx]int, n)	dep := make([]uint, n)	var dfs func(int, int) int	dfs = func(v, fa int) (size int) {		pa[v][0] = fa		dfn++		nodes[v].l = dfn		for _, w := range g[v] {			if w != fa {				dep[w] = dep[v] + 1				sz := dfs(w, v)				size += sz			}		}		size++		nodes[v].r = nodes[v].l + size		return	}	dfs(0, -1)	for i := 0; i+1 < mx; i++ {		for v := range pa {			if p := pa[v][i]; p != -1 {				pa[v][i+1] = pa[p][i]			} else {				pa[v][i+1] = -1			}		}	}	isAncestor := func(f, v int) bool { return nodes[f].l < nodes[v].l && nodes[v].l < nodes[f].r }	up := func(v int, d uint) int {		for k := dep[v] - d; k > 0; k &= k - 1 {			v = pa[v][bits.TrailingZeros(k)]		}		return v	} o:	for Fscan(in, &Q); Q > 0; Q-- {		Fscan(in, &k)		a := make([]int, k)		for i := range a {			Fscan(in, &a[i])			a[i]--			if dep[a[i]] < dep[a[0]] {				a[i], a[0] = a[0], a[i]			}		}		if k <= 2 {			Fprintln(out, "YES")			continue		}		p, q := a[0], a[1] // p 的深度最小		top := isAncestor(p, q)		if top {			aq = up(q, dep[p]+1) // aq 是 p 的儿子和 q 的祖先		}		for _, v := range a[2:] {			if top {				if isAncestor(q, v) { // v 在 q 下面					q = v				} else if !isAncestor(v, q) { // v 不在 p 和 q 之间					if isAncestor(aq, v) { // v 在 p 和 q 之间的分叉上						Fprintln(out, "NO")						continue o					}					p = v					top = false				}			} else if isAncestor(p, v) { // v 在 p 下面				p = v			} else if isAncestor(q, v) { // v 在 q 下面				q = v			} else if !isAncestor(v, p) && !isAncestor(v, q) { // v 不在 p 到 q 的路径上				Fprintln(out, "NO")				continue o			}		}		Fprintln(out, "YES")	}} //func main() { cf1702G2(os.Stdin, os.Stdout) }

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