Question

1443. Minimum Time to Collect All Apples in a Tree

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Description

Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. _Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming back to this vertex._

The edges of the undirected tree are given in the array edges, where edges[i] = [ai, bi] means that exists an edge connecting the vertices ai and bi. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not have any apple.

 

Example 1:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false]
Output: 8 
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.  

Example 2:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false]
Output: 6
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.  

Example 3:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,false,false,false,false,false]
Output: 0

 

Constraints:

• 1 <= n <= 105
• edges.length == n - 1
• edges[i].length == 2
• 0 <= ai < bi <= n - 1
• hasApple.length == n

Solutions

Solution 1

class Solution:
  def minTime(self, n: int, edges: List[List[int]], hasApple: List[bool]) -> int:
    def dfs(u, cost):
      if vis[u]:
        return 0
      vis[u] = True
      nxt_cost = 0
      for v in g[u]:
        nxt_cost += dfs(v, 2)
      if not hasApple[u] and nxt_cost == 0:
        return 0
      return cost + nxt_cost

    g = defaultdict(list)
    for u, v in edges:
      g[u].append(v)
      g[v].append(u)
    vis = [False] * n
    return dfs(0, 0)

class Solution {
  public int minTime(int n, int[][] edges, List<Boolean> hasApple) {
    boolean[] vis = new boolean[n];
    List<Integer>[] g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    for (int[] e : edges) {
      int u = e[0], v = e[1];
      g[u].add(v);
      g[v].add(u);
    }
    return dfs(0, 0, g, hasApple, vis);
  }

  private int dfs(int u, int cost, List<Integer>[] g, List<Boolean> hasApple, boolean[] vis) {
    if (vis[u]) {
      return 0;
    }
    vis[u] = true;
    int nxtCost = 0;
    for (int v : g[u]) {
      nxtCost += dfs(v, 2, g, hasApple, vis);
    }
    if (!hasApple.get(u) && nxtCost == 0) {
      return 0;
    }
    return cost + nxtCost;
  }
}

class Solution {
public:
  int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {
    vector<bool> vis(n);
    vector<vector<int>> g(n);
    for (auto& e : edges) {
      int u = e[0], v = e[1];
      g[u].push_back(v);
      g[v].push_back(u);
    }
    return dfs(0, 0, g, hasApple, vis);
  }

  int dfs(int u, int cost, vector<vector<int>>& g, vector<bool>& hasApple, vector<bool>& vis) {
    if (vis[u]) return 0;
    vis[u] = true;
    int nxt = 0;
    for (int& v : g[u]) nxt += dfs(v, 2, g, hasApple, vis);
    if (!hasApple[u] && !nxt) return 0;
    return cost + nxt;
  }
};

func minTime(n int, edges [][]int, hasApple []bool) int {
  vis := make([]bool, n)
  g := make([][]int, n)
  for _, e := range edges {
    u, v := e[0], e[1]
    g[u] = append(g[u], v)
    g[v] = append(g[v], u)
  }
  var dfs func(int, int) int
  dfs = func(u, cost int) int {
    if vis[u] {
      return 0
    }
    vis[u] = true
    nxt := 0
    for _, v := range g[u] {
      nxt += dfs(v, 2)
    }
    if !hasApple[u] && nxt == 0 {
      return 0
    }
    return cost + nxt
  }
  return dfs(0, 0)
}