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Problem solution · Go

Codeforces 758E — Broken Tree

Codeforces 758E — Broken Tree: a Go solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Depth-first search
Source
EndlessCheng Codeforces Go
Length
89 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

Depth-first search

For Codeforces 758E — Broken Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 89 lines of Go from the credited upstream file 758E.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 758E — Broken Tree · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // https://github.com/EndlessChengfunc cf758E(in io.Reader, _w io.Writer) {	out := bufio.NewWriter(_w)	defer out.Flush()	var n int	Fscan(in, &n)	type edge struct{ v, w, wt, p int }	es := make([]edge, n-1)	type nb struct{ to, i int }	g := make([][]nb, n+1)	for i := range es {		var v, w, wt, p int		Fscan(in, &v, &w, &wt, &p)		es[i] = edge{v, w, wt, p}		g[v] = append(g[v], nb{w, i})	} 	// 第一次 DFS：预处理每棵子树的 minSum（子树最小重量和）以及 extraDec（见下面注释）	a := make([]struct{ minSum, extraDec int }, n+1)	var dfs func(int) (int, int)	dfs = func(v int) (minSum, maxSum int) {		for _, e := range g[v] {			w := e.to			// 递归计算子树 w 的最小重量和 mn、最大重量和 mx			mn, mx := dfs(w)			p := es[e.i].p			// v-w 边的强度不能小于子树 w 的最小重量和			if mn < 0 || p < mn {				return -1, 0			}			wt := es[e.i].wt			// v-w 边的强度可以减少到子树 w 的最小重量和			minSum += max(wt-(p-mn), 1) + mn			// 子树 w 的最大重量和不能超过 v-w 边的强度			maxSum += wt + min(mx, p)			// 如果 v-w 这条边的强度 p < mx，那么 w 子树的最大重量和 mx 要额外减少 mx-p（减少到 p）			// 说【额外】是因为 mx 已经是 w 子树内部减重之后的最大重量和了			// 如果 p < mx，那么 mx 内部还要再减少 mx-p			a[w].extraDec = max(mx-p, 0)		}		a[v].minSum = minSum		// 最后返回子树 v 的最小重量和、最大重量和		return	}	minSum, _ := dfs(1)	if minSum < 0 {		Fprint(out, -1)		return	} 	// 第二次 DFS：减重	// 核心思想：优先减重最下面的边	// 如果不这样做，先减重上面的，那么由于上面的边强度变小，下面的边也得跟着减重，不如先减重下面的边优	dec := 0	var modify func(int)	modify = func(v int) {		for _, ew := range g[v] {			w := ew.to			// 递归之前，只累加需要减重的量，在递归之后处理减重，这样就可以保证下面的边先减重			dec += a[w].extraDec			// 处理 w 子树内部的减重			modify(w)			// 处理 v-w 这条边的减重			e := &es[ew.i]			// v-w 这条边，重量可以减到 1，强度可以减到子树 w 的最小重量和			d := min(e.wt-1, e.p-a[w].minSum, dec)			e.wt -= d			e.p -= d			dec -= d		}	}	modify(1) 	Fprintln(out, n)	for _, e := range es {		Fprintln(out, e.v, e.w, e.wt, e.p)	}} //func main() { cf758E(bufio.NewReader(os.Stdin), os.Stdout) }

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