CPMagical CP Library

Problem solution · Go

Codeforces 1106E — Lunar New Year and Red Envelopes

Codeforces 1106E — Lunar New Year and Red Envelopes: a Go solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Direct simulation
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

Direct simulation

For Codeforces 1106E — Lunar New Year and Red Envelopes, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 87 lines of Go from the credited upstream file 1106E.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 1106E — Lunar New Year and Red Envelopes · 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 dw6 struct{ d, w int }type pair6 struct {	dw6	i int}type mh6 []*pair6 func (h mh6) Len() int            { return len(h) }func (h mh6) Less(i, j int) bool  { a, b := h[i], h[j]; return a.w > b.w || a.w == b.w && a.d > b.d }func (h mh6) Swap(i, j int)       { h[i], h[j] = h[j], h[i]; h[i].i = i; h[j].i = j }func (h *mh6) Push(v interface{}) { *h = append(*h, v.(*pair6)) }func (h *mh6) Pop() interface{}   { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }func (h *mh6) push(e dw6) *pair6  { p := &pair6{e, len(*h)}; heap.Push(h, p); return p } func CF1106E(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	min := func(a, b int64) int64 {		if a < b {			return a		}		return b	} 	var n, m, k, s, t, d, w int	Fscan(in, &n, &m, &k)	es := make([][]dw6, n+1)	for ; k > 0; k-- {		Fscan(in, &s, &t, &d, &w)		es[s] = append(es[s], dw6{d, w})		if t < n {			es[t+1] = append(es[t+1], dw6{-d, w})		}	} 	next := make([]dw6, n+1)	h := mh6{}	ptr := map[dw6][]*pair6{}	for i, es := range es {		for _, e := range es {			if e.d > 0 {				ptr[e] = append(ptr[e], h.push(e))			} else {				e.d = -e.d				heap.Remove(&h, ptr[e][0].i)				ptr[e] = ptr[e][1:]			}		}		if len(h) > 0 {			next[i] = h[0].dw6		} else {			next[i].d = i		}	} 	ans := int64(1e18)	cur := make([]int64, n+2)	for i := range cur {		cur[i] = 1e18	}	cur[1] = 0	for k := 0; k <= m; k++ {		nxt := make([]int64, n+2)		for i := range nxt {			nxt[i] = 1e18		}		for i := 1; i <= n; i++ {			nxt[i+1] = min(nxt[i+1], cur[i])			cur[next[i].d+1] = min(cur[next[i].d+1], cur[i]+int64(next[i].w))		}		ans = min(ans, cur[n+1])		cur = nxt	}	Fprint(out, ans)} //func main() { CF1106E(os.Stdin, os.Stdout) }

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗