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Problem solution · Go

Codeforces 1927G — Paint Charges

Codeforces 1927G — Paint Charges: 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
74 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 1927G — Paint Charges, 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

  • 74 lines of Go from the credited upstream file 1927G.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 1927G — Paint Charges · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // https://space.bilibili.com/206214func cf1927G(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	var T, n int	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &n)		a := make([]int, n, n+1)		for i := range a {			Fscan(in, &a[i])		}		a = append(a, 1) 		dp := make([][][]int, n)		for i := range dp {			dp[i] = make([][]int, n+1)			for j := range dp[i] {				dp[i][j] = make([]int, n+1)				for k := range dp[i][j] {					dp[i][j][k] = -1				}			}		}		var f func(int, int, int) int		f = func(i, j, doneL int) int {			if doneL == 0 {				return 0			}			if i < 0 {				return 1e9			}			p := &dp[i][j][doneL]			if *p != -1 {				return *p			} 			// [doneL, n-1] 已完成			// 如果 j != n，则表示 [j+1, doneL-1] 未完成 			// 不选			res := f(i-1, j, doneL) 			// 向左			if j == n {				if i >= doneL-1 {					res = min(res, f(i-1, n, min(doneL, max(i-a[i]+1, 0)))+1)				} else {					res = min(res, f(i-1, i, doneL)+1) // 不连续				}			} else if i >= j-a[j] && i-a[i] < j-a[j] {				res = min(res, f(i-1, i, doneL)+1) // 仍然不连续，但是最左可以被更新			} 			// 向右			if i < doneL && i+a[i] >= doneL {				res = min(res, f(i-1, n, min(i, max(j-a[j]+1, 0)))+1)			} 			*p = res			return res		}		Fprintln(out, f(n-1, n, n))	}} //func main() { cf1927G(os.Stdin, os.Stdout) }

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