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Problem solution · Go

Codeforces 1753D — The Beach

Codeforces 1753D — The Beach: a Go solution using stack-based processing. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Stack-based processing
Source
EndlessCheng Codeforces Go
Length
98 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

Stack-based processing

For Codeforces 1753D — The Beach, the implementation keeps unresolved items in last-in, first-out order, often to match boundaries, parse structure, or maintain monotonic candidates.

  1. Define what every stack entry represents.
  2. Pop entries once the current item resolves or invalidates them.
  3. Push the current item with only the information later steps need.

Code notes

  • 98 lines of Go from the credited upstream file 1753D.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

If each item is pushed and popped at most once, the stack work is linear.

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 1753D — The Beach · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	"container/heap"	. "fmt"	"io") // https://space.bilibili.com/206214func CF1753D(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	const inf int64 = 1e18	dir4 := []struct{ x, y int }{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}	match := []struct{ x, y int }{'L': {0, 1}, 'R': {0, -1}, 'U': {1, 0}, 'D': {-1, 0}}	min := func(a, b int64) int64 {		if a > b {			return b		}		return a	} 	var n, m int	var p, q int64	Fscan(in, &n, &m, &p, &q)	g := make([]string, n)	dis := make([][]int64, n)	h := hp53{}	for i := range g {		Fscan(in, &g[i])		dis[i] = make([]int64, m)		for j, b := range g[i] {			if b == '.' {				dis[i][j] = 0				h.push(pd53{i, j, 0})			} else {				dis[i][j] = inf			}		}	}	for len(h) > 0 {		top := h.pop()		i, j := top.x, top.y		if top.dis > dis[i][j] {			continue		}		for _, d := range dir4 {			x, y := i+d.x, j+d.y			if 0 <= x && x < n && 0 <= y && y < m && g[x][y] != '.' && g[x][y] != '#' {				d := match[g[x][y]]				x += d.x				y += d.y				newD := dis[i][j]				if x != i && y != j {					newD += p				} else {					newD += q				}				if newD < dis[x][y] {					dis[x][y] = newD					h.push(pd53{x, y, newD})				}			}		}	}	ans := inf	for i, r := range dis {		for j, d := range r {			if j < m-1 {				ans = min(ans, d+r[j+1])			}			if i < n-1 {				ans = min(ans, d+dis[i+1][j])			}		}	}	if ans == inf {		ans = -1	}	Fprint(out, ans)} //func main() { CF1753D(os.Stdin, os.Stdout) } type pd53 struct {	x, y int	dis  int64}type hp53 []pd53 func (h hp53) Len() int              { return len(h) }func (h hp53) Less(i, j int) bool    { return h[i].dis < h[j].dis }func (h hp53) Swap(i, j int)         { h[i], h[j] = h[j], h[i] }func (h *hp53) Push(v interface{})   { *h = append(*h, v.(pd53)) }func (h *hp53) Pop() (v interface{}) { a := *h; *h, v = a[:len(a)-1], a[len(a)-1]; return }func (h *hp53) push(v pd53)          { heap.Push(h, v) }func (h *hp53) pop() pd53            { return heap.Pop(h).(pd53) }

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