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Problem solution · Go

Codeforces 1486E — Paired Payment

Codeforces 1486E — Paired Payment: 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
81 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 1486E — Paired Payment, 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

  • 81 lines of Go from the credited upstream file 1486E.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 1486E — Paired Payment · 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 pair86 struct{ v, d int }type hp86 []pair86 func (h hp86) Len() int              { return len(h) }func (h hp86) Less(i, j int) bool    { return h[i].d < h[j].d }func (h hp86) Swap(i, j int)         { h[i], h[j] = h[j], h[i] }func (h *hp86) Push(v interface{})   { *h = append(*h, v.(pair86)) }func (h *hp86) Pop() (v interface{}) { a := *h; *h, v = a[:len(a)-1], a[len(a)-1]; return }func (h *hp86) push(v pair86)        { heap.Push(h, v) }func (h *hp86) pop() pair86          { return heap.Pop(h).(pair86) } func CF1486E(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush()	const inf int = 1e9	type nb struct{ to, wt int } 	var n, m, v, w, wt int	Fscan(in, &n, &m)	g := make([][]nb, n)	for i := 0; i < m; i++ {		Fscan(in, &v, &w, &wt)		v--		w--		g[v] = append(g[v], nb{w, wt})		g[w] = append(g[w], nb{v, wt})	}	dist := make([]int, n)	mi := make([]int, n)	for i := range dist {		dist[i] = inf		mi[i] = inf	}	dist[0] = 0	q := hp86{{}}	for len(q) > 0 {		p := q.pop()		v := p.v		if p.d > dist[v] {			continue		}		for _, e := range g[v] {			w := e.to			// 优先队列出来的点肯定是先近后远			// 假设 v 比 v' 先出来			// 对于任意中间点 w			// 如果 v 可以用距离 d 和 w 做中转，那么没有必要再考虑 v' 用 >=d 的距离和 w 做中转了			if e.wt >= mi[w] {				continue			}			mi[w] = e.wt			for _, e2 := range g[w] {				u := e2.to				if newD := dist[v] + (e.wt+e2.wt)*(e.wt+e2.wt); newD < dist[u] {					dist[u] = newD					q.push(pair86{u, newD})				}			}		}	}	for _, v := range dist {		if v == inf {			v = -1		}		Fprint(out, v, " ")	}} //func main() { CF1486E(os.Stdin, os.Stdout) }

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