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困难
2300
第 220 场周赛 Q4
并查集
数组
双指针
排序

English Version

题目描述

给你一个 n 个点组成的无向图边集 edgeList ,其中 edgeList[i] = [ui, vi, disi] 表示点 ui 和点 vi 之间有一条长度为 disi 的边。请注意,两个点之间可能有 超过一条边 

给你一个查询数组queries ,其中 queries[j] = [pj, qj, limitj] ,你的任务是对于每个查询 queries[j] ,判断是否存在从 pj 到 qj 的路径,且这条路径上的每一条边都 严格小于 limitj 。

请你返回一个 布尔数组 answer ,其中 answer.length == queries.length ,当 queries[j] 的查询结果为 true 时, answer j 个值为 true ,否则为 false 。

 

示例 1:

输入:n = 3, edgeList = [[0,1,2],[1,2,4],[2,0,8],[1,0,16]], queries = [[0,1,2],[0,2,5]]
输出:[false,true]
解释:上图为给定的输入数据。注意到 0 和 1 之间有两条重边,分别为 2 和 16 。
对于第一个查询,0 和 1 之间没有小于 2 的边,所以我们返回 false 。
对于第二个查询,有一条路径(0 -> 1 -> 2)两条边都小于 5 ,所以这个查询我们返回 true 。

示例 2:

输入:n = 5, edgeList = [[0,1,10],[1,2,5],[2,3,9],[3,4,13]], queries = [[0,4,14],[1,4,13]]
输出:[true,false]
解释:上图为给定数据。

 

提示:

  • 2 <= n <= 105
  • 1 <= edgeList.length, queries.length <= 105
  • edgeList[i].length == 3
  • queries[j].length == 3
  • 0 <= ui, vi, pj, qj <= n - 1
  • ui != vi
  • pj != qj
  • 1 <= disi, limitj <= 109
  • 两个点之间可能有 多条 边。

解法

方法一:离线查询 + 并查集

根据题目要求,我们需要对每个查询 $queries[i]$ 进行判断,即判断当前查询的两个点 $a$$b$ 之间是否存在一条边权小于等于 $limit$ 的路径。

判断两点是否连通可以通过并查集来实现。另外,由于查询的顺序对结果没有影响,因此我们可以先将所有查询按照 $limit$ 从小到大排序,所有边也按照边权从小到大排序。

然后对于每个查询,我们从边权最小的边开始,将边权严格小于 $limit$ 的所有边加入并查集,接着利用并查集的查询操作判断两点是否连通即可。

时间复杂度 $O(m \times \log m + q \times \log q)$,其中 $m$$q$ 分别为边数和查询数。

Python3

class Solution:
    def distanceLimitedPathsExist(
        self, n: int, edgeList: List[List[int]], queries: List[List[int]]
    ) -> List[bool]:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(n))
        edgeList.sort(key=lambda x: x[2])
        j = 0
        ans = [False] * len(queries)
        for i, (a, b, limit) in sorted(enumerate(queries), key=lambda x: x[1][2]):
            while j < len(edgeList) and edgeList[j][2] < limit:
                u, v, _ = edgeList[j]
                p[find(u)] = find(v)
                j += 1
            ans[i] = find(a) == find(b)
        return ans

Java

class Solution {
    private int[] p;

    public boolean[] distanceLimitedPathsExist(int n, int[][] edgeList, int[][] queries) {
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        Arrays.sort(edgeList, (a, b) -> a[2] - b[2]);
        int m = queries.length;
        boolean[] ans = new boolean[m];
        Integer[] qid = new Integer[m];
        for (int i = 0; i < m; ++i) {
            qid[i] = i;
        }
        Arrays.sort(qid, (i, j) -> queries[i][2] - queries[j][2]);
        int j = 0;
        for (int i : qid) {
            int a = queries[i][0], b = queries[i][1], limit = queries[i][2];
            while (j < edgeList.length && edgeList[j][2] < limit) {
                int u = edgeList[j][0], v = edgeList[j][1];
                p[find(u)] = find(v);
                ++j;
            }
            ans[i] = find(a) == find(b);
        }
        return ans;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

C++

class Solution {
public:
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& queries) {
        vector<int> p(n);
        iota(p.begin(), p.end(), 0);
        sort(edgeList.begin(), edgeList.end(), [](auto& a, auto& b) { return a[2] < b[2]; });
        function<int(int)> find = [&](int x) -> int {
            if (p[x] != x) p[x] = find(p[x]);
            return p[x];
        };
        int m = queries.size();
        vector<bool> ans(m);
        vector<int> qid(m);
        iota(qid.begin(), qid.end(), 0);
        sort(qid.begin(), qid.end(), [&](int i, int j) { return queries[i][2] < queries[j][2]; });
        int j = 0;
        for (int i : qid) {
            int a = queries[i][0], b = queries[i][1], limit = queries[i][2];
            while (j < edgeList.size() && edgeList[j][2] < limit) {
                int u = edgeList[j][0], v = edgeList[j][1];
                p[find(u)] = find(v);
                ++j;
            }
            ans[i] = find(a) == find(b);
        }
        return ans;
    }
};

Go

func distanceLimitedPathsExist(n int, edgeList [][]int, queries [][]int) []bool {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	sort.Slice(edgeList, func(i, j int) bool { return edgeList[i][2] < edgeList[j][2] })
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	m := len(queries)
	qid := make([]int, m)
	ans := make([]bool, m)
	for i := range qid {
		qid[i] = i
	}
	sort.Slice(qid, func(i, j int) bool { return queries[qid[i]][2] < queries[qid[j]][2] })
	j := 0
	for _, i := range qid {
		a, b, limit := queries[i][0], queries[i][1], queries[i][2]
		for j < len(edgeList) && edgeList[j][2] < limit {
			u, v := edgeList[j][0], edgeList[j][1]
			p[find(u)] = find(v)
			j++
		}
		ans[i] = find(a) == find(b)
	}
	return ans
}

Rust

impl Solution {
    #[allow(dead_code)]
    pub fn distance_limited_paths_exist(
        n: i32,
        edge_list: Vec<Vec<i32>>,
        queries: Vec<Vec<i32>>,
    ) -> Vec<bool> {
        let mut disjoint_set: Vec<usize> = vec![0; n as usize];
        let mut ans_vec: Vec<bool> = vec![false; queries.len()];
        let mut q_vec: Vec<usize> = vec![0; queries.len()];

        // Initialize the set
        for i in 0..n {
            disjoint_set[i as usize] = i as usize;
        }

        // Initialize the q_vec
        for i in 0..queries.len() {
            q_vec[i] = i;
        }

        // Sort the q_vec based on the query limit, from the lowest to highest
        q_vec.sort_by(|i, j| queries[*i][2].cmp(&queries[*j][2]));

        // Sort the edge_list based on the edge weight, from the lowest to highest
        let mut edge_list = edge_list.clone();
        edge_list.sort_by(|i, j| i[2].cmp(&j[2]));

        let mut edge_idx: usize = 0;
        for q_idx in &q_vec {
            let s = queries[*q_idx][0] as usize;
            let d = queries[*q_idx][1] as usize;
            let limit = queries[*q_idx][2];
            // Construct the disjoint set
            while edge_idx < edge_list.len() && edge_list[edge_idx][2] < limit {
                Solution::union(
                    edge_list[edge_idx][0] as usize,
                    edge_list[edge_idx][1] as usize,
                    &mut disjoint_set,
                );
                edge_idx += 1;
            }
            // If the parents of s & d are the same, this query should be `true`
            // Otherwise, the current query is `false`
            ans_vec[*q_idx] = Solution::check_valid(s, d, &mut disjoint_set);
        }

        ans_vec
    }

    #[allow(dead_code)]
    pub fn find(x: usize, d_set: &mut Vec<usize>) -> usize {
        if d_set[x] != x {
            d_set[x] = Solution::find(d_set[x], d_set);
        }
        return d_set[x];
    }

    #[allow(dead_code)]
    pub fn union(s: usize, d: usize, d_set: &mut Vec<usize>) {
        let p_s = Solution::find(s, d_set);
        let p_d = Solution::find(d, d_set);
        d_set[p_s] = p_d;
    }

    #[allow(dead_code)]
    pub fn check_valid(s: usize, d: usize, d_set: &mut Vec<usize>) -> bool {
        let p_s = Solution::find(s, d_set);
        let p_d = Solution::find(d, d_set);
        p_s == p_d
    }
}

附并查集相关介绍以及常用模板:

并查集是一种树形的数据结构,顾名思义,它用于处理一些不交集的合并查询问题。 它支持两种操作:

  1. 查找(Find):确定某个元素处于哪个子集,单次操作时间复杂度 $O(\alpha(n))$
  2. 合并(Union):将两个子集合并成一个集合,单次操作时间复杂度 $O(\alpha(n))$

其中 $\alpha$ 为阿克曼函数的反函数,其增长极其缓慢,也就是说其单次操作的平均运行时间可以认为是一个很小的常数。

以下是并查集的常用模板,需要熟练掌握。其中:

  • n 表示节点数
  • p 存储每个点的父节点,初始时每个点的父节点都是自己
  • size 只有当节点是祖宗节点时才有意义,表示祖宗节点所在集合中,点的数量
  • find(x) 函数用于查找 $x$ 所在集合的祖宗节点
  • union(a, b) 函数用于合并 $a$$b$ 所在的集合

Python3

p = list(range(n))
size = [1] * n

def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]


def union(a, b):
    pa, pb = find(a), find(b)
    if pa == pb:
        return
    p[pa] = pb
    size[pb] += size[pa]

Java

int[] p = new int[n];
int[] size = new int[n];
for (int i = 0; i < n; ++i) {
    p[i] = i;
    size[i] = 1;
}

int find(int x) {
    if (p[x] != x) {
        // 路径压缩
        p[x] = find(p[x]);
    }
    return p[x];
}

void union(int a, int b) {
    int pa = find(a), pb = find(b);
    if (pa == pb) {
        return;
    }
    p[pa] = pb;
    size[pb] += size[pa];
}

C++

vector<int> p(n);
iota(p.begin(), p.end(), 0);
vector<int> size(n, 1);

int find(int x) {
    if (p[x] != x) {
        // 路径压缩
        p[x] = find(p[x]);
    }
    return p[x];
}

void unite(int a, int b) {
    int pa = find(a), pb = find(b);
    if (pa == pb) return;
    p[pa] = pb;
    size[pb] += size[pa];
}

Go

p := make([]int, n)
size := make([]int, n)
for i := range p {
    p[i] = i
    size[i] = 1
}

func find(x int) int {
    if p[x] != x {
        // 路径压缩
        p[x] = find(p[x])
    }
    return p[x]
}

func union(a, b int) {
    pa, pb := find(a), find(b)
    if pa == pb {
        return
    }
    p[pa] = pb
    size[pb] += size[pa]
}