水资源分配优化
难度:
标签:
题目描述
代码结果
运行时间: 93 ms, 内存: 22.3 MB
// Leetcode 1168: Optimize Water Distribution in a Village
// Problem: Given a number of houses and a number of wells, we need to find the minimum cost to supply water to all houses.
// Solution: This problem can be solved using Kruskal's algorithm for Minimum Spanning Tree (MST). We treat each house and well as nodes in a graph,
// and the pipes as edges with weights corresponding to the costs. We then find the MST of this graph to get the minimum cost.
import java.util.*;
import java.util.stream.*;
class Solution {
public int minCostToSupplyWater(int n, int[] wells, int[][] pipes) {
// Create a list to hold all the edges (pipes and wells)
List<int[]> edges = IntStream.range(0, wells.length)
.mapToObj(i -> new int[]{0, i + 1, wells[i]})
.collect(Collectors.toList());
// Add the given pipes as edges
edges.addAll(Arrays.asList(pipes));
// Sort edges by cost
edges.sort(Comparator.comparingInt(a -> a[2]));
// Create a union-find (disjoint-set) structure
int[] parent = IntStream.range(0, n + 1).toArray();
// Function to find the root of a node
int find(int x) {
if (parent[x] != x) {
parent[x] = find(parent[x]);
}
return parent[x];
}
// Function to union two sets
void union(int x, int y) {
parent[find(x)] = find(y);
}
final int[] totalCost = {0};
// Kruskal's algorithm to find MST
edges.stream()
.filter(edge -> find(edge[0]) != find(edge[1]))
.forEach(edge -> {
union(edge[0], edge[1]);
totalCost[0] += edge[2];
});
return totalCost[0];
}
}
解释
方法:
该题解采用了并查集数据结构来解决最小生成树问题,具体为Kruskal算法的变体。首先,将每个水井视为与每个房子直接相连的特殊节点,将其连接成本作为边的权重。将所有的管道和水井到房子的连接都看作边,并按照成本对这些边进行排序。接下来,使用并查集来逐一检查排序后的边,如果连接的两个节点不在同一集合中,则将这两个节点合并,并将边的成本加到总成本中。通过并查集操作,保证图中不会形成环,最终当所有房子都被连接到水源后(即并查集的根节点包括所有节点),算法结束。
时间复杂度:
O((m+n)log(m+n))
空间复杂度:
O(m+n)
代码细节讲解
🦆
在Kruskal算法中,为什么首先需要将所有边按成本进行排序?这一步骤对算法的正确性和效率有何影响?
▷🦆
并查集中的路径压缩技术具体是如何实现的?它如何影响并查集操作的效率?
▷🦆
题解中提到的‘阿克曼函数的反函数α(n)’在并查集的上下文中具体是如何工作的?为什么在实际应用中可以视为常数?
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