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Problem solution · Go

Codeforces 723F — st-Spanning Tree

Codeforces 723F — st-Spanning 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
97 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 723F — st-Spanning 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

  • 97 lines of Go from the credited upstream file 723F.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 723F — st-Spanning Tree · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF723F(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush() 	var n, m, v, w, s, t, ds, dt, sv, tv, both int	Fscan(in, &n, &m)	g := make([][]int, n+1)	for ; m > 0; m-- {		Fscan(in, &v, &w)		g[v] = append(g[v], w)		g[w] = append(g[w], v)	}	Fscan(in, &s, &t, &ds, &dt) 	ans := make([][2]int, 0, n-1)	vis := make([]bool, n+1)	var f func(int)	f = func(v int) {		vis[v] = true		for _, w := range g[v] {			if w == s {				sv = v			} else if w == t {				tv = v			} else if !vis[w] {				ans = append(ans, [2]int{v, w}) // 不与 s t 相邻的点，可以直接连边（DFS 树即生成树）				f(w)			}		}	}	type st struct{ s, t int }	nb := []st{}	for i := 1; i <= n; i++ {		if !vis[i] && i != s && i != t { // 对删去 s 和 t 的图 DFS			sv, tv = -1, -1			f(i)			// 若该连通分量只与 s t 其中一个相连，可以任取一相邻点直接连边			if sv < 0 {				ans = append(ans, [2]int{tv, t})				dt--			} else if tv < 0 {				ans = append(ans, [2]int{sv, s})				ds--			} else {				both++			}			nb = append(nb, st{sv, tv})		}	} 	// 此时 s 和 t 还未连通，s 和 t 还需要连边	if ds < 1 || dt < 1 || ds+dt <= both {		Fprint(out, "No")		return	} 	if both == 0 { // s 和 t 相邻		ans = append(ans, [2]int{s, t})	} else {		conn := false		for _, p := range nb {			if p.s >= 0 && p.t >= 0 {				if conn {					if ds > 0 {						ans = append(ans, [2]int{p.s, s})						ds--					} else {						ans = append(ans, [2]int{p.t, t})						dt--					}				} else {					conn = true					ans = append(ans, [2]int{p.s, s}, [2]int{p.t, t})					ds--					dt--				}			}		}	}	Fprintln(out, "Yes")	for _, p := range ans {		Fprintln(out, p[0], p[1])	}} //func main() { CF723F(os.Stdin, os.Stdout) }

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