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Problem solution · Go

Codeforces 914E — Palindromes in a Tree

Codeforces 914E — Palindromes in a 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
126 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 914E — Palindromes in a 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

  • 126 lines of Go from the credited upstream file 914E.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 914E — Palindromes in a Tree · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"math") // https://space.bilibili.com/206214func cf914E(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	var n int	var s string	Fscan(in, &n)	g := make([][]int, n)	for i := 1; i < n; i++ {		var v, w int		Fscan(in, &v, &w)		v--		w--		g[v] = append(g[v], w)		g[w] = append(g[w], v)	}	Fscan(in, &s) 	deleted := make([]bool, n)	size := make([]int, n)	var findCentroid func(int, int, int) (int, int, int)	findCentroid = func(v, fa, compSize int) (minSize, ct, faCt int) {		minSize = math.MaxInt		maxSubSize := 0		size[v] = 1		for _, w := range g[v] {			if w != fa && !deleted[w] {				minSizeW, ctW, faCtW := findCentroid(w, v, compSize)				if minSizeW < minSize {					minSize, ct, faCt = minSizeW, ctW, faCtW				}				maxSubSize = max(maxSubSize, size[w])				size[v] += size[w]			}		}		maxSubSize = max(maxSubSize, compSize-size[v])		if maxSubSize < minSize {			minSize, ct, faCt = maxSubSize, v, fa		}		return	} 	ans := make([]int, n)	cnt := [1 << 20]int{} 	// 更新连通块（子树）信息	var updateCC func(int, int, int, int)	updateCC = func(v, fa, delta, mask int) {		mask ^= 1 << (s[v] - 'a')		cnt[mask] += delta		for _, w := range g[v] {			if w != fa && !deleted[w] {				updateCC(w, v, delta, mask)			}		}	} 	// 计算「经过 v 向上，在重心拐弯，到其它子树」的路径信息	var calc func(int, int, int) int	calc = func(v, fa, mask int) int {		mask ^= 1 << (s[v] - 'a')		// 单独计算：从 v 出发的路径信息		res := cnt[mask]		for i := 1; i < len(cnt); i <<= 1 {			res += cnt[mask^i]		}		// 把 v 下面的也加上，这样最终算出的是经过 v 的路径信息		for _, w := range g[v] {			if w != fa && !deleted[w] {				res += calc(w, v, mask)			}		}		ans[v] += res		return res	} 	var dfs func(int, int, int)	dfs = func(v, fa, compSize int) {		_, ct, faCt := findCentroid(v, fa, compSize) 		updateCC(ct, -1, 1, 0)		// 单独计算：从 ct 出发的路径信息		res := cnt[0]		for i := 1; i < len(cnt); i <<= 1 {			res += cnt[i]		}		// 再加上经过 ct 的路径信息		for _, w := range g[ct] {			if deleted[w] {				continue			}			// 排除 w 子树后再计算			updateCC(w, ct, -1, 1<<(s[ct]-'a'))			res += calc(w, ct, 0)			updateCC(w, ct, 1, 1<<(s[ct]-'a'))		}		ans[ct] += res / 2 // v->w 和 w->v 算了两次		updateCC(ct, -1, -1, 0) 		deleted[ct] = true		for _, w := range g[ct] {			if !deleted[w] {				if w != faCt {					dfs(w, ct, size[w])				} else {					dfs(w, ct, compSize-size[ct])				}			}		}	}	dfs(0, -1, n)	for _, v := range ans {		Fprint(out, v+1, " ")	}} //func main() { cf914E(os.Stdin, os.Stdout) }

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