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Problem solution · Go

Codeforces 933A — A Twisty Movement

Codeforces 933A — A Twisty Movement: 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
79 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 933A — A Twisty Movement, 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

  • 79 lines of Go from the credited upstream file 933A.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 933A — A Twisty Movement · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // 思路1：由于求解的是最长的 11...22...11...22...，可以枚举中间位置，分成左右两部分，每部分都为 11...22...，用前缀和搞定// 这样复杂度是 O(n^2)// 思路2：题目本质是将数组划分成 4 部分（11... 22... 11... 22...），那么定义 dp[i][j] 表示前 i 个数组成了 j 个部分的最长值// 遍历一遍即可求出 dp[n][4]，复杂度为 O(n)// 参考 https://www.luogu.com.cn/problem/solution/CF933A// 思路2对应题目 https://www.acwing.com/problem/content/3552/ // EXTRA: 数据范围不止 [1,2] http://acm.hdu.edu.cn/showproblem.php?pid=6357 // github.com/EndlessCheng/codeforces-gofunc CF933A(_r io.Reader, out io.Writer) {	max := func(a, b int) int {		if a > b {			return a		}		return b	}	in := bufio.NewReader(_r)	var n int	Fscan(in, &n)	a := make([]int, n)	sum := make([][2]int, n+1)	for i := range a {		Fscan(in, &a[i])		sum[i+1][0] = sum[i][0]		sum[i+1][1] = sum[i][1]		sum[i+1][a[i]-1]++	}	ans := 0	for i := range a {		maxL := 0		for _, s := range sum[:i+1] {			maxL = max(maxL, s[0]+sum[i][1]-s[1])		}		maxR := 0		for _, s := range sum[i:] {			maxR = max(maxR, s[0]-sum[i][0]+sum[n][1]-s[1])		}		ans = max(ans, maxL+maxR)	}	Fprint(out, ans)} func CF933A2(_r io.Reader, out io.Writer) {	max := func(a, b int) int {		if a > b {			return a		}		return b	}	in := bufio.NewReader(_r)	var n, v int	Fscan(in, &n)	dp := [5]int{}	for ; n > 0; n-- {		if Fscan(in, &v); v == 1 {			dp[1]++			dp[2] = max(dp[1], dp[2])			dp[3] = max(dp[2], dp[3]+1)			dp[4] = max(dp[3], dp[4])		} else {			dp[2] = max(dp[1], dp[2]+1)			dp[3] = max(dp[2], dp[3])			dp[4] = max(dp[3], dp[4]+1)		}	}	Fprint(out, dp[4])} //func main() { CF933A(os.Stdin, os.Stdout) }

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