Question

2204. Distance to a Cycle in Undirected Graph

中文文档

Description

You are given a positive integer n representing the number of nodes in a connected undirected graph containing exactly one cycle. The nodes are numbered from 0 to n - 1 (inclusive).

You are also given a 2D integer array edges, where edges[i] = [node1i, node2i] denotes that there is a bidirectional edge connecting node1i and node2i in the graph.

The distance between two nodes a and b is defined to be the minimum number of edges that are needed to go from a to b.

Return _an integer array answer__ of size _n_, where _answer[i]_ is the minimum distance between the _ith_ node and any node in the cycle._

 

Example 1:

Input: n = 7, edges = [[1,2],[2,4],[4,3],[3,1],[0,1],[5,2],[6,5]]
Output: [1,0,0,0,0,1,2]
Explanation:
The nodes 1, 2, 3, and 4 form the cycle.
The distance from 0 to 1 is 1.
The distance from 1 to 1 is 0.
The distance from 2 to 2 is 0.
The distance from 3 to 3 is 0.
The distance from 4 to 4 is 0.
The distance from 5 to 2 is 1.
The distance from 6 to 2 is 2.

Example 2:

Input: n = 9, edges = [[0,1],[1,2],[0,2],[2,6],[6,7],[6,8],[0,3],[3,4],[3,5]]
Output: [0,0,0,1,2,2,1,2,2]
Explanation:
The nodes 0, 1, and 2 form the cycle.
The distance from 0 to 0 is 0.
The distance from 1 to 1 is 0.
The distance from 2 to 2 is 0.
The distance from 3 to 1 is 1.
The distance from 4 to 1 is 2.
The distance from 5 to 1 is 2.
The distance from 6 to 2 is 1.
The distance from 7 to 2 is 2.
The distance from 8 to 2 is 2.

 

Constraints:

• 3 <= n <= 105
• edges.length == n
• edges[i].length == 2
• 0 <= node1i, node2i <= n - 1
• node1i != node2i
• The graph is connected.
• The graph has exactly one cycle.
• There is at most one edge between any pair of vertices.

Solutions

Solution 1: Topological Sorting

We can first convert the edges in $edges$ into an adjacency list $g$, where $g[i]$ represents all adjacent nodes of node $i$, represented as a set.

Next, we delete nodes layer by layer from the outside to the inside until only a cycle remains. The specific method is as follows:

We first find all nodes with a degree of $1$ and remove these nodes from the graph. If after deletion, the degree of its adjacent node becomes $1$, then we add it to the queue $q$. During this process, we record all deleted nodes in order as $seq$; and we use an array $f$ to record the adjacent node of each node that is closer to the cycle, i.e., $f[i]$ represents the adjacent node of node $i$ that is closer to the cycle.

Finally, we initialize an answer array $ans$ of length $n$, where $ans[i]$ represents the minimum distance from node $i$ to any node in the cycle, initially $ans[i] = 0$. Then, we start traversing from the end of $seq$. For each node $i$, we can get the value of $ans[i]$ from its adjacent node $f[i]$, i.e., $ans[i] = ans[f[i]] + 1$.

After the traversal, return the answer array $ans$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes.

Similar problems:

• 2603. Collect Coins in a Tree

class Solution:
  def distanceToCycle(self, n: int, edges: List[List[int]]) -> List[int]:
    g = defaultdict(set)
    for a, b in edges:
      g[a].add(b)
      g[b].add(a)
    q = deque(i for i in range(n) if len(g[i]) == 1)
    f = [0] * n
    seq = []
    while q:
      i = q.popleft()
      seq.append(i)
      for j in g[i]:
        g[j].remove(i)
        f[i] = j
        if len(g[j]) == 1:
          q.append(j)
      g[i].clear()
    ans = [0] * n
    for i in seq[::-1]:
      ans[i] = ans[f[i]] + 1
    return ans

class Solution {
  public int[] distanceToCycle(int n, int[][] edges) {
    Set<Integer>[] g = new Set[n];
    Arrays.setAll(g, k -> new HashSet<>());
    for (var e : edges) {
      int a = e[0], b = e[1];
      g[a].add(b);
      g[b].add(a);
    }
    Deque<Integer> q = new ArrayDeque<>();
    for (int i = 0; i < n; ++i) {
      if (g[i].size() == 1) {
        q.offer(i);
      }
    }
    int[] f = new int[n];
    Deque<Integer> seq = new ArrayDeque<>();
    while (!q.isEmpty()) {
      int i = q.poll();
      seq.push(i);
      for (int j : g[i]) {
        g[j].remove(i);
        f[i] = j;
        if (g[j].size() == 1) {
          q.offer(j);
        }
      }
    }
    int[] ans = new int[n];
    while (!seq.isEmpty()) {
      int i = seq.pop();
      ans[i] = ans[f[i]] + 1;
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> distanceToCycle(int n, vector<vector<int>>& edges) {
    unordered_set<int> g[n];
    for (auto& e : edges) {
      int a = e[0], b = e[1];
      g[a].insert(b);
      g[b].insert(a);
    }
    queue<int> q;
    for (int i = 0; i < n; ++i) {
      if (g[i].size() == 1) {
        q.push(i);
      }
    }
    int f[n];
    int seq[n];
    int k = 0;
    while (!q.empty()) {
      int i = q.front();
      q.pop();
      seq[k++] = i;
      for (int j : g[i]) {
        g[j].erase(i);
        f[i] = j;
        if (g[j].size() == 1) {
          q.push(j);
        }
      }
      g[i].clear();
    }
    vector<int> ans(n);
    for (; k; --k) {
      int i = seq[k - 1];
      ans[i] = ans[f[i]] + 1;
    }
    return ans;
  }
};

func distanceToCycle(n int, edges [][]int) []int {
  g := make([]map[int]bool, n)
  for i := range g {
    g[i] = map[int]bool{}
  }
  for _, e := range edges {
    a, b := e[0], e[1]
    g[a][b] = true
    g[b][a] = true
  }
  q := []int{}
  for i := 0; i < n; i++ {
    if len(g[i]) == 1 {
      q = append(q, i)
    }
  }
  f := make([]int, n)
  seq := []int{}
  for len(q) > 0 {
    i := q[0]
    q = q[1:]
    seq = append(seq, i)
    for j := range g[i] {
      delete(g[j], i)
      f[i] = j
      if len(g[j]) == 1 {
        q = append(q, j)
      }
    }
    g[i] = map[int]bool{}
  }
  ans := make([]int, n)
  for k := len(seq) - 1; k >= 0; k-- {
    i := seq[k]
    ans[i] = ans[f[i]] + 1
  }
  return ans
}

function distanceToCycle(n: number, edges: number[][]): number[] {
  const g: Set<number>[] = new Array(n).fill(0).map(() => new Set<number>());
  for (const [a, b] of edges) {
    g[a].add(b);
    g[b].add(a);
  }
  const q: number[] = [];
  for (let i = 0; i < n; ++i) {
    if (g[i].size === 1) {
      q.push(i);
    }
  }
  const f: number[] = Array(n).fill(0);
  const seq: number[] = [];
  while (q.length) {
    const i = q.pop()!;
    seq.push(i);
    for (const j of g[i]) {
      g[j].delete(i);
      f[i] = j;
      if (g[j].size === 1) {
        q.push(j);
      }
    }
    g[i].clear();
  }
  const ans: number[] = Array(n).fill(0);
  while (seq.length) {
    const i = seq.pop()!;
    ans[i] = ans[f[i]] + 1;
  }
  return ans;
}