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Problem solution · Go

Codeforces 463D — Gargari and Permutations

Codeforces 463D — Gargari and Permutations: 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
42 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 463D — Gargari and Permutations, 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

  • 42 lines of Go from the credited upstream file 463D.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 463D — Gargari and Permutations · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // https://space.bilibili.com/206214func CF463D(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	var n, k, ans int	Fscan(in, &n, &k)	a := make([]int, n)	index := make([][]int, k)	for i := range index {		index[i] = make([]int, n+1)		for j := range a {			Fscan(in, &a[j])			index[i][a[j]] = j		}	} 	f := make([]int, n)	for i, x := range a { // 以最后一个排列 a 为基准	next:		for j, y := range a[:i] { // 枚举在 x 左边的数 y			for _, idx := range index {				if idx[y] > idx[x] { // 对于其余排列，y 的位置必须在 x 的左边					continue next				}			}			f[i] = max(f[i], f[j])		}		f[i]++		ans = max(ans, f[i])	}	Fprint(out, ans)} //func main() { CF463D(os.Stdin, os.Stdout) }

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