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Problem solution · Go

Codeforces 1511F — Chainword

Codeforces 1511F — Chainword: 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
115 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 1511F — Chainword, 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

  • 115 lines of Go from the credited upstream file 1511F.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 1511F — Chainword · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // https://github.com/EndlessChengconst mod11 = 998244353 type matrix11 [][]int func newMatrix11(n, m int) matrix11 {	a := make(matrix11, n)	for i := range a {		a[i] = make([]int, m)	}	return a} func (a matrix11) mul(b matrix11) matrix11 {	c := newMatrix11(len(a), len(b[0]))	for i, row := range a {		for k, x := range row {			if x == 0 {				continue			}			for j, y := range b[k] {				c[i][j] = (c[i][j] + x*y) % mod11			}		}	}	return c} func (a matrix11) powMul(n int, f0 matrix11) matrix11 {	res := f0	for ; n > 0; n /= 2 {		if n%2 > 0 {			res = a.mul(res)		}		a = a.mul(a)	}	return res} func cf1511F(in io.Reader, out io.Writer) {	var n, k int	var s string	Fscan(in, &n, &k) 	type node struct {		son [26]*node		end bool	}	root := &node{}	for range n {		Fscan(in, &s)		o := root		for _, b := range s {			b -= 'a'			if o.son[b] == nil {				o.son[b] = &node{}			}			o = o.son[b]		}		o.end = true	} 	m := newMatrix11(145, 145)	type pair struct{ p, q *node }	idx := map[pair]int{}	var dfs func(*node, *node) int	dfs = func(p, q *node) int {		t := pair{p, q}		if i, ok := idx[t]; ok {			return i		}		v := len(idx)		idx[t] = v		for i := range 26 {			a, b := p.son[i], q.son[i]			if a == nil || b == nil {				continue			}			m[dfs(a, b)][v] = 1			if a.end {				m[dfs(root, b)][v] = 1			}			if b.end {				m[dfs(root, a)][v]++			}			if a.end && b.end {				m[dfs(root, root)][v]++			}		}		return v	}	dfs(root, root) 	size := len(idx)	m = m[:size]	for i, row := range m {		m[i] = row[:size]	} 	f0 := newMatrix11(size, 1)	f0[0][0] = 1 	fn := m.powMul(k, f0)	Fprint(out, fn[0][0]%mod11)} //func main() { cf1511F(os.Stdin, os.Stdout) }

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