CPMagical CP Library

Problem solution · Go

Codeforces 1314D

Codeforces 1314D: 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
61 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 1314D, 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

  • 61 lines of Go from the credited upstream file 1314D.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 1314D · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"math/rand"	"time") // github.com/EndlessCheng/codeforces-gofunc CF1314D(_r io.Reader, out io.Writer) {	rand.Seed(time.Now().UnixNano())	in := bufio.NewReader(_r)	min := func(a, b int) int {		if a > b {			return b		}		return a	} 	var n, k int	Fscan(in, &n, &k)	dis := make([][]int, n)	for i := range dis {		dis[i] = make([]int, n)		for j := range dis[i] {			Fscan(in, &dis[i][j])		}	} 	ans := int(1e9)	dp := make([]int, n)	color := make([]int, n)	for t := 5000; t > 0; t-- {		dp[0] = 0		for i := 1; i < n; i++ {			dp[i] = 1e9			color[i] = rand.Intn(2)		}		for i := 0; i < k; i++ {			for v, c := range color {				if c != i&1 {					dp[v] = 1e9					for from, c := range color {						if c == i&1 {							// dp[i][v] 表示走了 i 条边，当前在节点 v 时的最小花费，那么答案就是 dp[k][0]							// 其中第一维可以压缩掉							dp[v] = min(dp[v], dp[from]+dis[from][v])						}					}				}			}		}		ans = min(ans, dp[0])	}	Fprint(out, ans)} //func main() { CF1314D(os.Stdin, os.Stdout) }

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗