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Problem solution · Go

Codeforces 1218D — Xor Spanning Tree

Codeforces 1218D — Xor Spanning Tree: a Go solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Direct simulation
Source
EndlessCheng Codeforces Go
Length
106 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

Direct simulation

For Codeforces 1218D — Xor Spanning Tree, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 106 lines of Go from the credited upstream file 1218D.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 1218D — Xor Spanning Tree · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // 本题需要用 int 存储计算，否则会 TLE// 更进一步的优化见 https://codeforces.com/contest/1218/submission/118704484 // github.com/EndlessCheng/codeforces-gofunc CF1218D(_r io.Reader, out io.Writer) {	in := bufio.NewReader(_r)	const mod = 1e9 + 7	const inv2 = (mod + 1) / 2	const mx = 1 << 17	fwt := func(a []int) {		for l, k := 2, 1; l <= mx; l, k = l<<1, k<<1 {			for i := 0; i < mx; i += l {				for j := 0; j < k; j++ {					a[i+j], a[i+j+k] = (a[i+j]+a[i+j+k])%mod, (a[i+j]-a[i+j+k])%mod				}			}		}	}	ifwt := func(a []int) {		for l, k := 2, 1; l <= mx; l, k = l<<1, k<<1 {			for i := 0; i < mx; i += l {				for j := 0; j < k; j++ {					a[i+j], a[i+j+k] = int(int64(a[i+j]+a[i+j+k])*inv2%mod), int(int64(a[i+j]-a[i+j+k])*inv2%mod)				}			}		}	} 	var n, m, v, w, wt, xor int	Fscan(in, &n, &m)	type nb struct{ to, wt int }	g := make([][]nb, n+1)	for ; m > 0; m-- {		Fscan(in, &v, &w, &wt)		g[v] = append(g[v], nb{w, wt})		g[w] = append(g[w], nb{v, wt})		xor ^= wt	} 	// 仙人掌找环	cnt := [][]int{}	s := []nb{{1, 0}}	vis := make([]int8, n+1)	var f func(int, int)	f = func(v, fa int) {		vis[v] = 1		for _, e := range g[v] {			if w := e.to; vis[w] == 0 {				s = append(s, e)				f(w, v)			} else if w != fa && vis[w] == 1 {				c := make([]int, mx)				for i := len(s) - 1; s[i].to != w; i-- {					c[s[i].wt]++				}				c[e.wt]++				cnt = append(cnt, c)			}		}		vis[v] = 2		s = s[:len(s)-1]	}	f(1, 0) 	has := make([]int, mx)	for i, c := range cnt[0] {		if c != 0 {			has[i] = 1		}	}	for _, c := range cnt {		fwt(c)	}	for _, c := range cnt[1:] {		fwt(has)		for j, v := range c {			cnt[0][j] = int(int64(cnt[0][j]) * int64(v) % mod)			has[j] = int(int64(has[j]) * int64(v) % mod)		}		ifwt(has)		for i, v := range has {			if v != 0 {				has[i] = 1			}		}	}	ifwt(cnt[0]) 	for i := 0; ; i++ {		if has[xor^i] != 0 {			Fprint(out, i, (cnt[0][xor^i]+mod)%mod)			break		}	}} //func main() { CF1218D(os.Stdin, os.Stdout) }

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