https://leetcode.com/problems/optimize-water-distribution-in-a-village/
There are n houses in a village. We want to supply water for all the houses by building wells and laying pipes.
For each house i, we can either build a well inside it directly with cost wells[i], or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes, where each pipes[i] = [house1, house2, cost] represents the cost to connect house1 and house2 together using a pipe. Connections are bidirectional.
Find the minimum total cost to supply water to all houses.
Example 1:
Input: n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]]
Output: 3
Explanation:
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.
Constraints:
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]
example 1 pic:
题意，在每个城市打井需要一定的花费，也可以用其他城市的井水，城市之间建立连接管道需要一定的花费，怎么样安排可以花费最少的前灌溉所有城市。
这是一道连通所有点的最短路径/最小生成树问题，把城市看成图中的点，管道连接城市看成是连接两个点之间的边。这里打井的花费是直接在点上，而且并不是所有
点之间都有边连接，为了方便，我们可以假想一个点（root）0，这里自身点的花费可以与 0 连接，花费可以是 0-i 之间的花费。这样我们就可以构建一个连通图包含所有的点和边。
那在一个连通图中求最短路径/最小生成树的问题.
参考延伸阅读中，维基百科针对这类题给出的几种解法。
解题步骤：
- 创建
POJO EdgeCost(node1, node2, cost) - 节点1 和 节点2 连接边的花费。 - 假想一个
root点0，构建图 - 连通所有节点和
0，[0,i] - i 是节点 [1,n]，0-1是节点0和1的边，边的值是节点i上打井的花费wells[i]; - 把打井花费和城市连接点转换成图的节点和边。
- 对图的边的值排序（从小到大）
- 遍历图的边，判断两个节点有没有连通 （
Union-Find），- 已连通就跳过，继续访问下一条边
- 没有连通，记录花费，连通节点
- 若所有节点已连通，求得的最小路径即为最小花费，返回
- 对于每次
union, 节点数n-1, 如果n==0说明所有节点都已连通，可以提前退出，不需要继续访问剩余的边。
这里用加权Union-Find 判断两个节点是否连通，和连通未连通的节点。
举例：n = 5, wells=[1,2,2,3,2], pipes=[[1,2,1],[2,3,1],[4,5,7]]
如图：
从图中可以看到，最后所有的节点都是连通的。
- 时间复杂度:
O(ElogE) - E 是图的边的个数 - 空间复杂度:
O(E)
一个图最多有
n(n-1)/2 - n 是图中节点个数条边 （完全连通图）
- 构建图，得出所有边
- 对所有边排序
- 遍历所有的边（从小到大）
- 对于每条边，检查是否已经连通，若没有连通，加上边上的值，连通两个节点。若已连通，跳过。
Java code
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