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Problem solution · Go

Codeforces 590C — Three States

Codeforces 590C — Three States: a Go solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Breadth-first search
Source
EndlessCheng Codeforces Go
Length
97 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

Breadth-first search

For Codeforces 590C — Three States, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 97 lines of Go from the credited upstream file 590C.go.
  • The implementation visibly relies on sequence storage, work queue.
  • No explicit loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 590C — Three States · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gotype pair90 struct{ x, y int }type deque90 struct{ l, r []pair90 } func (q deque90) empty() bool     { return len(q.l) == 0 && len(q.r) == 0 }func (q *deque90) pushL(v pair90) { q.l = append(q.l, v) }func (q *deque90) pushR(v pair90) { q.r = append(q.r, v) }func (q *deque90) popL() (v pair90) {	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 CF590C(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	dir4 := []pair90{{-1, 0}, {1, 0}, {0, -1}, {0, 1}} 	var n, m int	Fscan(in, &n, &m)	g := make([][]byte, n)	for i := range g {		Fscan(in, &g[i])	}	bfs := func(tp byte) [][]int {		dis := make([][]int, n)		for i := range dis {			dis[i] = make([]int, m)			for j := range dis[i] {				dis[i][j] = 1e8			}		}		q := &deque90{}		for i, r := range g {			for j, b := range r {				if b == tp {					dis[i][j] = 0					q.pushL(pair90{i, j})				}			}		}		for !q.empty() {			p := q.popL()			for _, d := range dir4 {				if x, y := p.x+d.x, p.y+d.y; 0 <= x && x < n && 0 <= y && y < m && g[x][y] != '#' {					d := 0					if g[x][y] == '.' {						d = 1					}					if newD := dis[p.x][p.y] + d; newD < dis[x][y] {						dis[x][y] = newD						if d == 0 {							q.pushL(pair90{x, y})						} else {							q.pushR(pair90{x, y})						}					}				}			}		}		return dis	}	d1 := bfs('1')	d2 := bfs('2')	d3 := bfs('3')	ans := int(1e7)	for i, r := range g {		for j, b := range r {			if b != '#' {				d := d1[i][j] + d2[i][j] + d3[i][j]				if b == '.' {					d -= 2				}				if d < ans {					ans = d				}			}		}	}	if ans == 1e7 {		ans = -1	}	Fprint(out, ans)} //func main() { CF590C(os.Stdin, os.Stdout) }

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