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Problem solution · Go

Codeforces 1156D — 0-1-Tree

Codeforces 1156D — 0-1-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
62 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 1156D — 0-1-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

  • 62 lines of Go from the credited upstream file 1156D.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 1156D — 0-1-Tree · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // github.com/EndlessCheng/codeforces-gofunc Sol1156D(reader io.Reader, writer io.Writer) {	var find func([]int, int) int	find = func(fa []int, i int) int {		if fa[i] != i {			fa[i] = find(fa, fa[i])		}		return fa[i]	}	merge := func(fa []int, size []int, from, to int) {		from, to = find(fa, from), find(fa, to)		if from != to {			fa[from] = to			size[to] += size[from]		}	} 	in := bufio.NewReader(reader)	out := bufio.NewWriter(writer)	defer out.Flush() 	var n int	Fscan(in, &n)	fa := [2][]int{}	size := [2][]int{}	for i := range fa {		fa[i] = make([]int, n)		for j := range fa[i] {			fa[i][j] = j		}		size[i] = make([]int, n)		for j := range size[i] {			size[i][j] = 1		}	}	for m := n - 1; m > 0; m-- {		var v, w, c int		Fscan(in, &v, &w, &c)		merge(fa[c], size[c], v-1, w-1)	} 	ans := int64(0)	for i := range fa[0] {		// 考察顶点 v，假设通过 0-边与其相连的点一共有 x 个（包括 v），通过 1-边与其相连的有 y 个（包括 v），		// 那么以 x 个点为起点，以 y 个点为终点的路径条数 x*y-1 即为顶点 v 对答案的贡献（-1 是因为 v 到 v 不算）。		ans += int64(size[0][find(fa[0], i)])*int64(size[1][find(fa[1], i)]) - 1	}	Fprint(out, ans)} //func main() {//	Sol1156D(os.Stdin, os.Stdout)//}

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