CPMagical CP Library

Problem solution · Go

Codeforces 453C — Little Pony and Summer Sun Celebration

Codeforces 453C — Little Pony and Summer Sun Celebration: 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
77 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 453C — Little Pony and Summer Sun Celebration, 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

  • 77 lines of Go from the credited upstream file 453C.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 453C — Little Pony and Summer Sun Celebration · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF453C(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush() 	var n, m, v, w int	Fscan(in, &n, &m)	g := make([][]int, n)	for ; m > 0; m-- {		Fscan(in, &v, &w)		v--		w--		g[v] = append(g[v], w)		g[w] = append(g[w], v)	}	odd := make([]bool, n)	for i := range odd {		Fscan(in, &odd[i])	} 	path := []int{}	vis := make([]bool, n)	var f func(int, int)	f = func(v, fa int) {		vis[v] = true		path = append(path, v)		odd[v] = !odd[v]		var son int		for _, w := range g[v] {			if !vis[w] {				f(w, v)				path = append(path, v)				odd[v] = !odd[v]				son = w			}		}		if odd[v] {			if fa < 0 { // 必定有 son				path = append(path, son, v, son)			} else {				path = append(path, fa, v)				odd[fa] = !odd[fa]			}			odd[v] = false		}	}	for i, b := range odd {		if b {			f(i, -1)			break		}	} 	for i, b := range odd {		if b && !vis[i] { // 还有没访问的 odd			Fprint(out, -1)			return		}	} 	Fprintln(out, len(path))	for _, v := range path {		Fprint(out, v+1, " ")	}} //func main() { CF453C(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 ↗