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Problem solution · Go

Codeforces 219C — Color Stripe

Codeforces 219C — Color Stripe: 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
56 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 219C — Color Stripe, 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

  • 56 lines of Go from the credited upstream file 219C.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 219C — Color Stripe · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // 另一种做法是 k=2 时答案肯定是 ABAB... 与 BABA... 中的一个；// k=3 时若当前字母与左侧相邻字母不同，则修改其为与左右相邻字母均不同的字母 // github.com/EndlessCheng/codeforces-gofunc CF219C(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush() 	var n, k, miJ int	var s string	Fscan(in, &n, &k, &s)	dp := make([]int, k)	fa := make([][26]int, n)	for i, b := range s {		b := int(b - 'A')		dp2 := make([]int, k)		for j := range dp2 {			dp2[j] = 1e9			add := 0			if j != b {				add = 1			}			for k, v := range dp {				if k != j && v+add < dp2[j] {					dp2[j] = v + add					fa[i][j] = k				}			}		}		dp = dp2	}	for j, v := range dp {		if v < dp[miJ] {			miJ = j		}	}	Fprintln(out, dp[miJ])	ans := make([]byte, n)	for i, j := n-1, miJ; i >= 0; i-- {		ans[i] = 'A' + byte(j)		j = fa[i][j]	}	Fprintf(out, "%s", ans)} //func main() { CF219C(os.Stdin, os.Stdout) }

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