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Problem solution · Go

Codeforces 626F — Group Projects

Codeforces 626F — Group Projects: 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
60 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 626F — Group Projects, 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

  • 60 lines of Go from the credited upstream file 626F.go.
  • The implementation visibly relies on sequence storage, cached states.
  • 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 626F — Group Projects · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"sort") // https://space.bilibili.com/206214func CF626F(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	const mod = 1_000_000_007	var n, k int	Fscan(in, &n, &k)	a := make([]int, n, n+1)	for i := range a {		Fscan(in, &a[i])	}	sort.Ints(a)	a = append(a, 0) // 避免越界 	memo := [200][101][1001]int{}	for i := range memo {		for j := range memo[i] {			for k := range memo[i][j] {				memo[i][j][k] = -1			}		}	}	var dfs func(int, int, int) int64	dfs = func(i, groups, leftK int) (res int64) {		if leftK < 0 || groups > i+1 { // groups > i+1 说明剩余数字不够组成最小值			return		}		if i < 0 {			if groups == 0 {				return 1			}			return		}		p := &memo[i][groups][leftK]		if *p != -1 {			return int64(*p)		}		leftK -= (a[i+1] - a[i]) * groups		res = dfs(i-1, groups+1, leftK) // a[i] 作为最大值		res += dfs(i-1, groups, leftK) * int64(groups+1) // 不参与最大最小：从 groups 中选一个组   这里 +1 是只有一个数的组的方案数		if groups > 0 {			res += dfs(i-1, groups-1, leftK) * int64(groups) // a[i] 作为最小值：从 groups 中选一个组		}		res %= mod		*p = int(res) // 记忆化		return	}	Fprint(out, dfs(n-1, 0, k))} //func main() { CF626F(os.Stdin, os.Stdout) }

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