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Problem solution · Go

Codeforces 1473E — Minimum Path

Codeforces 1473E — Minimum Path: 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
87 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 1473E — Minimum Path, 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

  • 87 lines of Go from the credited upstream file 1473E.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 1473E — Minimum Path · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	"container/heap"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gotype pair73 struct {	v int	d int64}type hp73 []pair73 func (h hp73) Len() int              { return len(h) }func (h hp73) Less(i, j int) bool    { return h[i].d < h[j].d }func (h hp73) Swap(i, j int)         { h[i], h[j] = h[j], h[i] }func (h *hp73) Push(v interface{})   { *h = append(*h, v.(pair73)) }func (h *hp73) Pop() (v interface{}) { a := *h; *h, v = a[:len(a)-1], a[len(a)-1]; return }func (h *hp73) push(v pair73)        { heap.Push(h, v) }func (h *hp73) pop() pair73          { return heap.Pop(h).(pair73) } func CF1473E(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	min := func(a, b int64) int64 {		if a < b {			return a		}		return b	} 	var n, m, v, w, wt int	Fscan(in, &n, &m)	type nb struct{ to, wt int }	g := make([][]nb, n)	for ; m > 0; m-- {		Fscan(in, &v, &w, &wt)		v--		w--		g[v] = append(g[v], nb{w, wt})		g[w] = append(g[w], nb{v, wt})	} 	const inf int64 = 1e18	dis := make([]int64, n*4)	for i := range dis {		dis[i] = inf	}	dis[0] = 0	h := hp73{{}}	for len(h) > 0 {		pd := h.pop()		x := pd.v		if dis[x] < pd.d {			continue		}		// 分成四层，分别表示原图、不算最大值、额外再算一遍最小值、不算最大值且额外再算一遍最小值		// 从第一层到第四层就是要求的最短路		// 即便当前的边不是最小或最大，后面转移的时候也会覆盖掉		for _, e := range g[x>>2] {			w, wt := e.to, int64(e.wt)			y := w<<2 | x&3			if newD := dis[x] + wt; newD < dis[y] {				dis[y] = newD				h.push(pair73{y, newD})			}			if newD := dis[x] + wt*2; x&1 == 0 && newD < dis[y|1] {				dis[y|1] = newD				h.push(pair73{y | 1, newD})			}			if newD := dis[x]; x>>1&1 == 0 && newD < dis[y|2] {				dis[y|2] = newD				h.push(pair73{y | 2, newD})			}		}	}	for i := 1; i < n; i++ {		Fprint(out, min(dis[i<<2], dis[i<<2|3]), " ")	}} //func main() { CF1473E(os.Stdin, os.Stdout) }

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