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Problem solution · Go

Codeforces 732F — Tourist Reform

Codeforces 732F — Tourist Reform: 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
122 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 732F — Tourist Reform, 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

  • 122 lines of Go from the credited upstream file 732F.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 732F — Tourist Reform · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF732F(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	min := func(a, b int) int {		if a > b {			return b		}		return a	} 	var n, m, v, w int	Fscan(in, &n, &m)	type nb struct{ to, eid int }	g := make([][]nb, n)	type edge struct{ v, w int }	es := make([]edge, m)	for i := range es {		Fscan(in, &v, &w)		v--		w--		g[v] = append(g[v], nb{w, i})		g[w] = append(g[w], nb{v, i})		es[i] = edge{v, w}	} 	isBridge := make([]bool, m)	dfn := make([]int, n)	ts := 0	var f func(int, int) int	f = func(v, fa int) int {		ts++		dfn[v] = ts		lowV := ts		for _, e := range g[v] {			if w := e.to; dfn[w] == 0 {				lowW := f(w, v)				if lowW > dfn[v] {					isBridge[e.eid] = true				}				lowV = min(lowV, lowW)			} else if w != fa {				lowV = min(lowV, dfn[w])			}		}		return lowV	}	for v, ts := range dfn {		if ts == 0 {			f(v, -1)		}	} 	bccIDs := make([]int, n)	idCnt := 0	vs := []int{}	vis := make([]bool, m)	var f2 func(int)	f2 = func(v int) {		bccIDs[v] = idCnt		vs = append(vs, v)		for _, e := range g[v] {			if w, i := e.to, e.eid; !isBridge[i] {				if !vis[i] { // 避免重复定向					vis[i] = true					es[i] = edge{v, w} // 定向：BCC 内				}				if bccIDs[w] == 0 {					f2(w)				}			}		}	}	rt, maxBCC := 0, 0	for v, id := range bccIDs {		if id == 0 {			idCnt++			vs = []int{}			f2(v)			if len(vs) > maxBCC {				rt, maxBCC = idCnt, len(vs)			}		}	} 	g2 := make([][]nb, idCnt)	for i, b := range isBridge {		if b {			v, w := bccIDs[es[i].v]-1, bccIDs[es[i].w]-1			g2[v] = append(g2[v], nb{w, i})			g2[w] = append(g2[w], nb{v, i})		}	}	var f3 func(v, fa int)	f3 = func(v, fa int) {		for _, e2 := range g2[v] {			if w, i := e2.to, e2.eid; w != fa {				if e := es[i]; bccIDs[e.w]-1 == w { // 方向和输入相反					es[i] = edge{e.w, e.v} // 定向：缩点后的 bridge tree 上				}				f3(w, v)			}		}	}	f3(rt-1, -1)	Fprintln(out, maxBCC)	for _, e := range es {		Fprintln(out, e.v+1, e.w+1)	}} //func main() { CF732F(os.Stdin, os.Stdout) }

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