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Problem solution · Go

Codeforces 1263F — Economic Difficulties

Codeforces 1263F — Economic Difficulties: 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
77 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 1263F — Economic Difficulties, 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

  • 77 lines of Go from the credited upstream file 1263F.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 1263F — Economic Difficulties · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc CF1263F(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	min := func(a, b int) int {		if a < b {			return a		}		return b	}	max := func(a, b int) int {		if a > b {			return a		}		return b	} 	var n, m, v int	Fscan(in, &n)	maxDel := make([][]int, n+1)	for i := range maxDel {		maxDel[i] = make([]int, n+1)	}	f := func() {		Fscan(in, &m)		g := make([][]int, m+1)		for w := 2; w <= m; w++ {			Fscan(in, &v)			g[v] = append(g[v], w)		}		tar := make([]int, m+1)		for i := 1; i <= n; i++ {			Fscan(in, &v)			tar[v] = i		}		var f func(int) (sz, l, r int)		f = func(v int) (edgeSize, l, r int) {			if v > 1 {				edgeSize = 1 // 到父节点的边			}			l = 1e9			if tar[v] > 0 {				l, r = tar[v], tar[v]			}			for _, w := range g[v] {				sz, ll, rr := f(w)				edgeSize += sz				l = min(l, ll)				r = max(r, rr)			}			// 直接求区间。因为电缆没有交叉，且设备编号是连续的（见题目中的 Formally, for each tree exists a depth-first search ... 这段）			maxDel[l][r] = max(maxDel[l][r], edgeSize)			return		}		f(1)	}	f()	f() 	dp := make([]int, n+1)	for i := range dp {		for j, dv := range dp[:i] {			dp[i] = max(dp[i], dv+maxDel[j+1][i])		}	}	Fprint(out, dp[n])} //func main() { CF1263F(os.Stdin, os.Stdout) }

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