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Problem solution · Go

Codeforces 375D — Tree and Queries

Codeforces 375D — Tree and Queries: 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
110 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 375D — Tree and Queries, 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

  • 110 lines of Go from the credited upstream file 375D.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 375D — Tree and Queries · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // https://space.bilibili.com/206214func cf375D(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush() 	var n, m, v, w, k, dfn int	Fscan(in, &n, &m)	a := make([]int, n)	for i := range a {		Fscan(in, &a[i])	}	g := make([][]int, n)	for i := 1; i < n; i++ {		Fscan(in, &v, &w)		v--		w--		g[v] = append(g[v], w)		g[w] = append(g[w], v)	}	type pair struct{ i, k int }	qs := make([][]pair, n)	for i := 0; i < m; i++ {		Fscan(in, &v, &k)		if k <= n {			qs[v-1] = append(qs[v-1], pair{i, k})		}	} 	nodes := make([]struct{ l, r, hson int }, n) // [l,r)	nodeVals := make([]int, 0, n)	var build func(int, int) int	build = func(v, fa int) int {		nodes[v].l = dfn		dfn++		nodeVals = append(nodeVals, a[v])		size, hsz, hson := 1, 0, -1		for _, w := range g[v] {			if w != fa {				sz := build(w, v)				size += sz				if sz > hsz {					hsz, hson = sz, w				}			}		}		nodes[v].r = nodes[v].l + size		nodes[v].hson = hson		return size	}	build(0, -1) 	ans := make([]int, m)	cnt := [1e5 + 1]int{}	cc := make([]int, n+1)	var f func(int, int)	f = func(v, fa int) {		hson := nodes[v].hson		for _, w := range g[v] {			if w == fa || w == hson {				continue			}			f(w, v)			// 恢复现场，这样下一棵子树不会受到影响			for _, x := range nodeVals[nodes[w].l:nodes[w].r] {				cc[cnt[x]]--				cnt[x]--			}		}		if hson >= 0 {			f(hson, v)			// 此时重儿子的数据已经添加		} 		// 添加根节点的数据		cnt[a[v]]++		cc[cnt[a[v]]]++		// 添加非重儿子的数据		for _, w := range g[v] {			if w == fa || w == hson {				continue			}			for _, x := range nodeVals[nodes[w].l:nodes[w].r] {				cnt[x]++				cc[cnt[x]]++			}		} 		// 子树 v 的所有数据添加完毕，回答询问		for _, q := range qs[v] {			ans[q.i] = cc[q.k]		}	}	f(0, -1) 	for _, v := range ans {		Fprintln(out, v)	}} //func main() { cf375D(os.Stdin, os.Stdout) }

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