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Problem solution · Go

Codeforces 808E — Selling Souvenirs

Codeforces 808E — Selling Souvenirs: a Go solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Dynamic programming
Source
EndlessCheng Codeforces Go
Length
64 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

Dynamic programming

For Codeforces 808E — Selling Souvenirs, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 64 lines of Go from the credited upstream file 808E.go.
  • The implementation visibly relies on sequence storage, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 808E — Selling Souvenirs · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"sort") // github.com/EndlessCheng/codeforces-gofunc CF808E(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	max := func(a, b int64) int64 {		if b > a {			return b		}		return a	} 	var n, maxW, w, v int	Fscan(in, &n, &maxW)	a := [3]sort.IntSlice{}	for ; n > 0; n-- {		Fscan(in, &w, &v)		a[w-1] = append(a[w-1], v)	}	for _, b := range a {		sort.Sort(sort.Reverse(b))	} 	// 枚举 3，然后把两个 1 看成一个 2，贪心取	// 这种做法也可以用 DP 来实现，dp[i] 表示重量 i 能取到的最大价值，以及该最大价值用了多少 1 和 2	type pair struct {		x, y int		res  int64	}	dp := make([]pair, maxW+1)	for i := 1; i <= maxW; i++ {		dp[i] = dp[i-1]		if dp[i].x < len(a[0]) {			dp[i].res += int64(a[0][dp[i].x])			dp[i].x++		}		if i > 1 && dp[i-2].y < len(a[1]) && dp[i-2].res+int64(a[1][dp[i-2].y]) > dp[i].res {			dp[i] = dp[i-2]			dp[i].res += int64(a[1][dp[i-2].y])			dp[i].y++		}	} 	ans := dp[maxW].res	s3 := int64(0)	for i, v := range a[2] {		if (i+1)*3 > maxW {			break		}		s3 += int64(v)		ans = max(ans, s3+dp[maxW-(i+1)*3].res)	}	Fprint(out, ans)} //func main() { CF808E(os.Stdin, os.Stdout) }

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