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Problem solution · Go

Codeforces 2026F — Bermart Ice Cream

Codeforces 2026F — Bermart Ice Cream: a Go solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Depth-first search
Source
EndlessCheng Codeforces Go
Length
144 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

Depth-first search

For Codeforces 2026F — Bermart Ice Cream, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 144 lines of Go from the credited upstream file 2026F.go.
  • The implementation visibly relies on sequence storage, work queue.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 2026F — Bermart Ice Cream · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // https://github.com/EndlessChengvar mem26 [90002][2001]int32 // todo 最多需要多少空间？var cur26 = 1 type tuple26 struct {	w, v, fi int} type stack26 []tuple26 func (st stack26) res(w int) int32 {	return mem26[st[len(st)-1].fi][w]} func (st *stack26) push(w, v int) {	cur26++	f := mem26[cur26][:]	copy(f, mem26[(*st)[len(*st)-1].fi][:])	for i := len(f) - 1; i >= w; i-- {		f[i] = max(f[i], f[i-w]+int32(v))	}	*st = append(*st, tuple26{w, v, cur26})} func (st *stack26) pop() (w, v int) {	n := len(*st) - 1	w, v = (*st)[n].w, (*st)[n].v	*st = (*st)[:n]	return} func (st stack26) empty() bool {	return len(st) == 1} type deque26 struct{ l, r stack26 } func (q *deque26) rebalance() {	if q.r.empty() {		q.l, q.r = q.r, q.l		defer func() { q.l, q.r = q.r, q.l }()	}	m := len(q.r) / 2	for i := m; i > 0; i-- {		q.l.push(q.r[i].w, q.r[i].v)	}	t := q.r[m+1:]	q.r = q.r[:1]	for _, p := range t {		q.r.push(p.w, p.v)	}} func (q deque26) res(w int) (mx int32) {	for i := range w + 1 {		mx = max(mx, q.l.res(i)+q.r.res(w-i))	}	return} func (q *deque26) pushFront(w, v int) {	q.l.push(w, v)} func (q *deque26) pushBack(w, v int) {	q.r.push(w, v)} func (q *deque26) popFront() (w, v int) {	if q.l.empty() {		q.rebalance()	}	return q.l.pop()} func (q *deque26) popBack() {	if q.r.empty() {		q.rebalance()	}	q.r.pop()} func cf2026F(in io.Reader, _w io.Writer) {	out := bufio.NewWriter(_w)	defer out.Flush()	var q, nodeId, op, x, p, t int	Fscanln(in, &q)	type edge struct{ to, op, p, t, i int }	g := make([][]edge, q+1)	pos := make([]int, q+1)	store := 1 	for i := range q {		Fscanln(in, &op, &x, &p, &t)		nodeId++		v := pos[x]		g[v] = append(g[v], edge{nodeId, op, p, t, i})		if op == 1 {			store++			pos[store] = nodeId		} else {			pos[x] = nodeId		}	} 	ans := make([]int32, q)	dq := deque26{stack26{{}}, stack26{{fi: 1}}}	var dfs func(int)	dfs = func(v int) {		for _, e := range g[v] {			if e.op == 2 {				dq.pushBack(e.p, e.t)			} else if e.op == 3 {				e.p, e.t = dq.popFront()			} else if e.op == 4 {				ans[e.i] = dq.res(e.p) + 1			}			dfs(e.to)			if e.op == 2 {				dq.popBack()			} else if e.op == 3 {				dq.pushFront(e.p, e.t)			}		}	}	dfs(0) 	for _, v := range ans {		if v > 0 {			Fprintln(out, v-1)		}	}} //func main() { cf2026F(bufio.NewReader(os.Stdin), os.Stdout) }

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