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Problem solution · Go

Codeforces 1700E — Serega the Pirate

Codeforces 1700E — Serega the Pirate: 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
89 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 1700E — Serega the Pirate, 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

  • 89 lines of Go from the credited upstream file 1700E.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 1700E — Serega the Pirate · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io") // https://github.com/EndlessChengfunc cf1700E(in io.Reader, out io.Writer) {	var n, m int	Fscan(in, &n, &m)	a := make([][]int32, n)	for i := range a {		a[i] = make([]int32, m)		for j := range a[i] {			Fscan(in, &a[i][j])		}	} 	// 好格子：自己是 1，或者存在一个小于自己的邻居	ok := func(i, j int) bool {		return a[i][j] == 1 ||			j > 0 && a[i][j-1] < a[i][j] ||			j+1 < m && a[i][j+1] < a[i][j] ||			i > 0 && a[i-1][j] < a[i][j] ||			i+1 < n && a[i+1][j] < a[i][j]	}	// 判断 (i,j)，以及 (i,j) 的邻居，是否都是好格子	ok2 := func(i, j int) bool {		return ok(i, j) &&			(j == 0 || ok(i, j-1)) &&			(j+1 == m || ok(i, j+1)) &&			(i == 0 || ok(i-1, j)) &&			(i+1 == n || ok(i+1, j))	} 	// 收集坏格子	type pair struct{ i, j int }	badPos := []pair{}	for i := range n {		for j := range m {			if !ok(i, j) {				badPos = append(badPos, pair{i, j})			}		}	}	if len(badPos) == 0 {		Fprint(out, 0)		return	} 	ans := map[pair]struct{}{}	// 除了交换 (bi,bj)，也可以通过交换 (bi,bj) 的邻居，使 (bi,bj) 变成一个好格子	// 只需检查至多 5 个位置，因为 (bi,bj) 必须变成好格子	bi, bj := badPos[0].i, badPos[0].j	for _, p := range []pair{{bi, bj}, {bi, bj - 1}, {bi, bj + 1}, {bi - 1, bj}, {bi + 1, bj}} {		if p.i < 0 || p.i == n || p.j < 0 || p.j == m {			continue		}		for i := range n {			for j := range m {				// 交换其他所有点				a[p.i][p.j], a[i][j] = a[i][j], a[p.i][p.j]				// 交换离坏格子很远的点，必然是无效交换，所以先检查是否有坏格子仍然是坏格子				for _, q := range badPos {					if !ok(q.i, q.j) {						goto o					}				}				// 有效交换！进一步检查受到影响的 10 个点是否正常				if ok2(p.i, p.j) && ok2(i, j) {					// 注意去重					ans[pair{min(p.i*m+p.j, i*m+j), max(p.i*m+p.j, i*m+j)}] = struct{}{}				}			o:				a[p.i][p.j], a[i][j] = a[i][j], a[p.i][p.j]			}		}	} 	if len(ans) > 0 {		Fprintln(out, 1, len(ans))	} else {		Fprint(out, 2)	}} //func main() { cf1700E(bufio.NewReader(os.Stdin), os.Stdout) }

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