村庄内有n户人家，我们可以通过挖井或者建造水管向每家供水。
对于每户人家i，我们可以通过花费 wells[i] 直接在其房内挖水井，或者通过水管连接到其他的水井。每两户住户间铺设水管的费用通过 pipes 数组表示。 pipes[i] = [house1, house2, cost] 表示住户1到住户2间铺设水管的费用为cost。
请求出所有住户都能通水的最小花费。
示例1：
输入: n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]]
输出: 3
解释:
The image shows the costs of connecting houses using pipes.
The best strategy is to build a well in the first house with cost 1 and connect the other houses to it with cost 2 so the total cost is 3.
提示：
1 <= n <= 10000
wells.length == n
0 <= wells[i] <= 10^5
1 <= pipes.length <= 10000
1 <= pipes[i][0], pipes[i][1] <= n
0 <= pipes[i][2] <= 10^5
pipes[i][0] != pipes[i][1]
class OptimizeWaterDistribution {
public int minCostToSupplyWater(int n, int[] wells, int[][] pipes) {
List<EdgeCost> costs = new ArrayList<>();
for (int i = 1; i <= n; i++) {
costs.add(new EdgeCost(0, i, wells[i - 1]));
}
for (int[] p : pipes) {
costs.add(new EdgeCost(p[0], p[1], p[2]));
}
Collections.sort(costs);
int minCosts = 0;
UnionFind uf = new UnionFind(n);
for (EdgeCost edge : costs) {
int rootX = uf.find(edge.node1);
int rootY = uf.find(edge.node2);
if (rootX == rootY) continue;
minCosts += edge.cost;
uf.union(edge.node1, edge.node2);
// for each union, we connnect one node
n--;
// if all nodes already connected, terminate early
if (n == 0) {
return minCosts;
}
}
return minCosts;
}
class EdgeCost implements Comparable<EdgeCost> {
int node1;
int node2;
int cost;
public EdgeCost(int node1, int node2, int cost) {
this.node1 = node1;
this.node2 = node2;
this.cost = cost;
}
@Override
public int compareTo(EdgeCost o) {
return this.cost - o.cost;
}
}
class UnionFind {
int[] parent;
int[] rank;
public UnionFind(int n) {
parent = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
}
rank = new int[n + 1];
}
public int find(int x) {
return x == parent[x] ? x : find(parent[x]);
}
public void union(int x, int y) {
int px = find(x);
int py = find(y);
if (px == py) return;
if (rank[px] >= rank[py]) {
parent[py] = px;
rank[px] += rank[py];
} else {
parent[px] = py;
rank[py] += rank[px];
}
}
}
}
Pythong3 code
class Solution:
def minCostToSupplyWater(self, n: int, wells: List[int], pipes: List[List[int]]) -> int:
union_find = {i: i for i in range(n + 1)}
def find(x):
return x if x == union_find[x] else find(union_find[x])
def union(x, y):
px = find(x)
py = find(y)
union_find[px] = py
graph_wells = [[cost, 0, i] for i, cost in enumerate(wells, 1)]
graph_pipes = [[cost, i, j] for i, j, cost in pipes]
min_costs = 0
for cost, x, y in sorted(graph_wells + graph_pipes):
if find(x) == find(y):
continue
union(x, y)
min_costs += cost
n -= 1
if n == 0:
return min_costs
延伸阅读
大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库：https://github.com/azl397985856/leetcode 。 目前已经 37K star 啦。 大家也可以关注我的公众号《力扣加加》带你啃下算法这块硬骨头。
