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Problem solution · Go

Codeforces 1442C — Graph Transpositions

Codeforces 1442C — Graph Transpositions: 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
110 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 1442C — Graph Transpositions, 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

  • 110 lines of Go from the credited upstream file 1442C.go.
  • The implementation visibly relies on sequence storage, work queue.
  • 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 1442C — Graph Transpositions · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gotype deque42 struct{ l, r []int } func (q deque42) empty() bool  { return len(q.l) == 0 && len(q.r) == 0 }func (q *deque42) pushL(v int) { q.l = append(q.l, v) }func (q *deque42) pushR(v int) { q.r = append(q.r, v) }func (q *deque42) popL() (v int) {	if len(q.l) > 0 {		q.l, v = q.l[:len(q.l)-1], q.l[len(q.l)-1]	} else {		v, q.r = q.r[0], q.r[1:]	}	return} func CF1442C(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	const mod = 998244353	pow := func(x int64, n int) (res int64) {		res = 1		for ; n > 0; n >>= 1 {			if n&1 > 0 {				res = res * x % mod			}			x = x * x % mod		}		return	} 	var n, m, v, w int	Fscan(in, &n, &m)	type nb struct{ to, wt int }	g := make([][]nb, n+1)	for ; m > 0; m-- {		Fscan(in, &v, &w)		g[v] = append(g[v], nb{w, 0})		g[w] = append(g[w], nb{v, 1})	} 	minTr := make([]int, n+1)	for i := range minTr {		minTr[i] = 1e9	}	minTr[1] = 0	q := &deque42{}	q.pushL(1)	for !q.empty() {		v := q.popL()		for _, e := range g[v] {			w, d := e.to, minTr[v]&1^e.wt // 根据 minTr[v] 决定实际方向			if newD := minTr[v] + d; newD < minTr[w] {				minTr[w] = newD				if d == 0 {					q.pushL(w)				} else {					q.pushR(w)				}			}		}	} 	dis := make([][20]int, n+1)	for i := range dis {		for j := range dis[i] {			dis[i][j] = 1e9		}	}	dis[1][0] = 0	type pair struct{ v, lv int }	q2 := []pair{{1, 0}}	for len(q2) > 0 {		p := q2[0]		q2 = q2[1:]		v := p.v		for _, e := range g[v] {			w, wt := e.to, p.lv&1^e.wt // 根据 p.lv 决定实际方向			// 通过减去 minTr 来得到映射到 dis 上的层数			if lv := p.lv + wt - minTr[w]; lv < 20 {				if newD := dis[v][p.lv-minTr[v]] + 1; newD < dis[w][lv] {					dis[w][lv] = newD					q2 = append(q2, pair{w, p.lv + wt})				}			}		}	} 	minT, minD := int(1e9), 0	for i, d := range dis[n] {		// t 为实际层数		if t := i + minTr[n]; minT > 19 || t > 19 {			if t < minT {				minT, minD = t, d			}		} else if 1<<t+d < 1<<minT+minD {			minT, minD = t, d		}	}	Fprint(out, (int64(minD)+pow(2, minT)-1)%mod)} //func main() { CF1442C(os.Stdin, os.Stdout) }

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