2021-08-25：给定数组father大小为N，表示一共有N个节点，father[i] = j 表示点i的父亲是点j， father表示的树一定是一棵树而不是森林，queries是二维数组，大小为

Posted 2021-09-11 福大大架构师每日一题

2021-08-25：给定数组father大小为N，表示一共有N个节点，father[i] = j 表示点i的父亲是点j， father表示的树一定是一棵树而不是森林，queries是二维数组，大小为M*2，每一个长度为2的数组都表示一条查询，[4,9], 表示想查询4和9之间的最低公共祖先…，[3,7], 表示想查询3和7之间的最低公共祖先…，tree和queries里面的所有值，都一定在0~N-1之间。返回一个数组ans，大小为M，ans[i]表示第i条查询的答案。

福大大 答案2021-08-25：

树链剖分。

代码用golang编写。代码如下：

package main

import "fmt"

func main() {
    father := []int{2, 3, 3, 3, 2, 1}
    queries := [][]int{
        []int{2, 0},
        []int{0, 4},
        []int{4, 5},
        []int{1, 0},
        []int{2, 2},
    }
    ret := query3(father, queries)
    fmt.Println(ret)
}

// 在线查询最优解 -> 树链剖分
// 空间复杂度O(N), 支持在线查询，单次查询时间复杂度O(logN)
// 如果有M次查询，时间复杂度O(N + M * logN)
func query3(father []int, queries [][]int) []int {
    tc := NewTreeChain(father)
    M := len(queries)
    ans := make([]int, M)
    for i := 0; i < M; i++ {
        // x y ?
        // x x x
        if queries[i][0] == queries[i][1] {
            ans[i] = queries[i][0]
        } else {
            ans[i] = tc.lca(queries[i][0], queries[i][1])
        }
    }
    return ans
}

type TreeChain struct {
    n    int
    h    int
    tree [][]int
    fa   []int
    dep  []int
    son  []int
    siz  []int
    top  []int
}

func NewTreeChain(father []int) *TreeChain {
    ans := &TreeChain{}
    ans.initTree(father)
    ans.dfs1(ans.h, 0)
    ans.dfs2(ans.h, ans.h)
    return ans
}

func (this *TreeChain) initTree(father []int) {
    this.n = len(father) + 1
    this.tree = make([][]int, this.n)
    this.fa = make([]int, this.n)
    this.dep = make([]int, this.n)
    this.son = make([]int, this.n)
    this.siz = make([]int, this.n)
    this.top = make([]int, this.n)
    this.n--
    cnum := make([]int, this.n)
    for i := 0; i < this.n; i++ {
        if father[i] == i {
            this.h = i + 1
        } else {
            cnum[father[i]]++
        }
    }
    this.tree[0] = make([]int, 0)
    for i := 0; i < this.n; i++ {
        this.tree[i+1] = make([]int, cnum[i])
    }
    for i := 0; i < this.n; i++ {
        if i+1 != this.h {
            cnum[father[i]]--
            this.tree[father[i]+1][cnum[father[i]]] = i + 1
        }
    }
}

func (this *TreeChain) dfs1(u int, f int) {
    this.fa[u] = f
    this.dep[u] = this.dep[f] + 1
    this.siz[u] = 1
    maxSize := -1
    for _, v := range this.tree[u] {
        this.dfs1(v, u)
        this.siz[u] += this.siz[v]
        if this.siz[v] > maxSize {
            maxSize = this.siz[v]
            this.son[u] = v
        }
    }
}

func (this *TreeChain) dfs2(u int, t int) {
    this.top[u] = t
    if this.son[u] != 0 {
        this.dfs2(this.son[u], t)
        for _, v := range this.tree[u] {
            if v != this.son[u] {
                this.dfs2(v, v)
            }
        }
    }
}

func (this *TreeChain) lca(a int, b int) int {
    a++
    b++
    for this.top[a] != this.top[b] {
        if this.dep[this.top[a]] > this.dep[this.top[b]] {
            a = this.fa[this.top[a]]
        } else {
            b = this.fa[this.top[b]]
        }
    }

    return twoSelectOne(this.dep[a] < this.dep[b], a, b) - 1
}

func twoSelectOne(c bool, a int, b int) int {
    if c {
        return a
    } else {
        return b
    }
}

执行结果如下：

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左神java代码