牛客题解 | 寻宝

题目

题目链接

解题思路

这是一个最小生成树问题，需要找到连接所有空地的最小生成树，并且要求最长的边最小。这种问题可以使用Kruskal算法的变体来解决。

关键点：

1. 使用Kruskal算法构建最小生成树
2. 按照边的长度排序
3. 使用并查集维护连通性
4. 找到满足条件的最小最大边

算法步骤：

1. 对所有边按长度排序
2. 使用二分查找最小的最大边长度
3. 检查是否能连通所有点
4. 返回满足条件的最小值

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代码

#include <bits/stdc++.h>
using namespace std;class UnionFind {
private:vector<int> parent;vector<int> rank;int count;public:UnionFind(int n) : parent(n), rank(n, 0), count(n) {for (int i = 0; i < n; i++) {parent[i] = i;}}int find(int x) {if (parent[x] != x) {parent[x] = find(parent[x]);}return parent[x];}void unite(int x, int y) {int rootX = find(x);int rootY = find(y);if (rootX != rootY) {if (rank[rootX] < rank[rootY]) {swap(rootX, rootY);}parent[rootY] = rootX;if (rank[rootX] == rank[rootY]) {rank[rootX]++;}count--;}}bool isConnected(int x, int y) {return find(x) == find(y);}int getCount() {return count;}
};class Solution {
public:int findMinMaxEdge(int n, vector<vector<int>>& edges) {// 按边长排序sort(edges.begin(), edges.end(), [](const vector<int>& a, const vector<int>& b) {return a[2] < b[2];});int left = 0, right = edges.back()[2];int result = right;while (left <= right) {int mid = left + (right - left) / 2;// 检查是否能用不超过mid长度的边连通所有点if (canConnect(n, edges, mid)) {result = mid;right = mid - 1;} else {left = mid + 1;}}return result;}private:bool canConnect(int n, const vector<vector<int>>& edges, int maxLen) {UnionFind uf(n + 1);// 只使用长度不超过maxLen的边for (const auto& edge : edges) {if (edge[2] > maxLen) break;uf.unite(edge[0], edge[1]);}// 检查是否所有点都连通（除了0）int root = uf.find(1);for (int i = 2; i <= n; i++) {if (uf.find(i) != root) return false;}return true;}
};int main() {ios::sync_with_stdio(false);cin.tie(nullptr);int n, m;cin >> n >> m;vector<vector<int>> edges(m, vector<int>(3));for (int i = 0; i < m; i++) {cin >> edges[i][0] >> edges[i][1] >> edges[i][2];}Solution solution;cout << solution.findMinMaxEdge(n, edges) << endl;return 0;
}

import java.util.*;class UnionFind {private int[] parent;private int[] rank;private int count;public UnionFind(int n) {parent = new int[n];rank = new int[n];count = n;for (int i = 0; i < n; i++) {parent[i] = i;}}public int find(int x) {if (parent[x] != x) {parent[x] = find(parent[x]);}return parent[x];}public void unite(int x, int y) {int rootX = find(x);int rootY = find(y);if (rootX != rootY) {if (rank[rootX] < rank[rootY]) {int temp = rootX;rootX = rootY;rootY = temp;}parent[rootY] = rootX;if (rank[rootX] == rank[rootY]) {rank[rootX]++;}count--;}}public boolean isConnected(int x, int y) {return find(x) == find(y);}public int getCount() {return count;}
}class Solution {private boolean canConnect(int n, int[][] edges, int maxLen) {UnionFind uf = new UnionFind(n + 1);// 只使用长度不超过maxLen的边for (int[] edge : edges) {if (edge[2] > maxLen) break;uf.unite(edge[0], edge[1]);}// 检查是否所有点都连通int root = uf.find(1);for (int i = 2; i <= n; i++) {if (uf.find(i) != root) return false;}return true;}public int findMinMaxEdge(int n, int[][] edges) {// 按边长排序Arrays.sort(edges, (a, b) -> Integer.compare(a[2], b[2]));int left = 0, right = edges[edges.length - 1][2];int result = right;while (left <= right) {int mid = left + (right - left) / 2;if (canConnect(n, edges, mid)) {result = mid;right = mid - 1;} else {left = mid + 1;}}return result;}
}public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();int[][] edges = new int[m][3];for (int i = 0; i < m; i++) {edges[i][0] = sc.nextInt();edges[i][1] = sc.nextInt();edges[i][2] = sc.nextInt();}Solution solution = new Solution();System.out.println(solution.findMinMaxEdge(n, edges));sc.close();}
}

class UnionFind:def __init__(self, n):self.parent = list(range(n))self.rank = [0] * nself.count = ndef find(self, x):if self.parent[x] != x:self.parent[x] = self.find(self.parent[x])return self.parent[x]def unite(self, x, y):root_x = self.find(x)root_y = self.find(y)if root_x != root_y:if self.rank[root_x] < self.rank[root_y]:root_x, root_y = root_y, root_xself.parent[root_y] = root_xif self.rank[root_x] == self.rank[root_y]:self.rank[root_x] += 1self.count -= 1def is_connected(self, x, y):return self.find(x) == self.find(y)def get_count(self):return self.countclass Solution:def can_connect(self, n: int, edges: list, max_len: int) -> bool:uf = UnionFind(n + 1)# 只使用长度不超过max_len的边for edge in edges:if edge[2] > max_len:breakuf.unite(edge[0], edge[1])# 检查是否所有点都连通root = uf.find(1)return all(uf.find(i) == root for i in range(2, n + 1))def find_min_max_edge(self, n: int, edges: list) -> int:# 按边长排序edges.sort(key=lambda x: x[2])left, right = 0, edges[-1][2]result = rightwhile left <= right:mid = left + (right - left) // 2if self.can_connect(n, edges, mid):result = midright = mid - 1else:left = mid + 1return resultdef main():n, m = map(int, input().split())edges = []for _ in range(m):p, q, k = map(int, input().split())edges.append([p, q, k])solution = Solution()print(solution.find_min_max_edge(n, edges))if __name__ == "__main__":main()

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算法及复杂度

• 算法：Kruskal算法 + 二分查找
• 时间复杂度：\(\mathcal{O(M \log M + M \log K)}\)，其中 \(M\) 是边数，\(K\) 是最大边长
• 空间复杂度：\(\mathcal{O(N)}\)，用于并查集数据结构