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Problem solution · Go

Codeforces 1453E — Dog Snacks

Codeforces 1453E — Dog Snacks: a Go solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Direct simulation
Source
EndlessCheng Codeforces Go
Length
68 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

Direct simulation

For Codeforces 1453E — Dog Snacks, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 68 lines of Go from the credited upstream file 1453E.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 1453E — Dog Snacks · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF1453E(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	max := func(a, b int) int {		if a > b {			return a		}		return b	} 	var T, n, v, w int	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &n)		g := make([][]int, n)		for i := 1; i < n; i++ {			Fscan(in, &v, &w)			v--			w--			g[v] = append(g[v], w)			g[w] = append(g[w], v)		}		ans := 0		var f func(v, fa int) int		f = func(v, fa int) int {			mi, mx := -1, 0			for _, w := range g[v] {				if w != fa {					k := f(w, v)					if mi < 0 || k < mi {						mi = k					}					mx = max(mx, k)				}			}			if len(g[v]) > 2 {				ans = max(ans, mx+2) // 儿子之间的移动			}			return mi + 1 // 从最小的儿子返回		}		mx, mx2 := 0, 0		for _, w := range g[0] {			k := f(w, 0)			if k > mx {				mx, mx2 = k, mx			} else if k > mx2 {				mx2 = k			}		}		ans = max(ans, mx+1) // 特殊处理根节点：从最深的儿子返回根节点		if len(g[0]) > 1 {			ans = max(ans, mx2+2) // 从非最深的儿子到其他儿子的移动		}		Fprintln(out, ans)	}} //func main() { CF1453E(os.Stdin, os.Stdout) }

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