CPMagical CP Library

Problem solution · Go

Codeforces 1584F — Strange LCS

Codeforces 1584F — Strange LCS: 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
85 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 1584F — Strange LCS, 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

  • 85 lines of Go from the credited upstream file 1584F.go.
  • The implementation visibly relies on sequence storage, cached states.
  • 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 1584F — Strange LCS · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // https://space.bilibili.com/206214func cf1584F(in io.Reader, out io.Writer) {	var T, n int	var s string	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &n)		pos := [123][10][]int{}		for i := 0; i < n; i++ {			pos[0][i] = []int{-1} // 假定在 LCS 前面还有个字符 \0，下标为 -1			Fscan(in, &s)			for j, b := range s {				pos[b][i] = append(pos[b][i], j) // 记录字母 b 在字符串 s[i] 中的出现位置 j			}		} 		memo := make([][123]int, 1<<n)		for i := range memo {			for j := range memo[i] {				memo[i][j] = -1			}		}		type pair struct{ mask int; c byte }		from := make([][123]pair, 1<<n) // 记录转移来源		var dfs func(int, byte) int		dfs = func(mask int, c byte) (res int) {			p := &memo[mask][c]			if *p != -1 {				return *p			}			var frm pair			// 枚举 LCS 的下一个字母 ch			// 要求：ch 在所有字符串中的下标 > c 在对应字符串中的下标			// 如果有两个 ch 都满足要求，优先取左边的，对应下面代码中的 p[0] > cur			for ch := byte('A'); ch <= 'z'; {				mask2 := 0				for i, p := range pos[ch][:n] {					if p == nil {						goto nxt					}					cur := pos[c][i][mask>>i&1] // 当前字母 c 的下标					// p[0] 或者 p[1] 是下一个字母 ch 的下标					if p[0] > cur {						// 0					} else if len(p) > 1 && p[1] > cur {						mask2 |= 1 << i					} else {						goto nxt					}				}				if r := dfs(mask2, ch); r > res {					res = r					frm.mask = mask2 // 记录转移来源					frm.c = ch				}			nxt:				if ch == 'Z' {					ch = 'a'				} else {					ch++				}			}			from[mask][c] = frm			res++			*p = res			return		}		Fprintln(out, dfs(0, 0)-1) 		lcs := []byte{}		for p := from[0][0]; p.c > 0; p = from[p.mask][p.c] {			lcs = append(lcs, p.c)		}		Fprintf(out, "%s\n", lcs)	}} //func main() { cf1584F(bufio.NewReader(os.Stdin), os.Stdout) }

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗