Optimize Water Distribution in a Village
Description
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 - 1]
(note the -1
due to 0-indexing), 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[j] = [house1j, house2j, costj]
represents the cost to connect house1j
and house2j
together using a pipe. Connections are bidirectional, and there could be multiple valid connections between the same two houses with different costs.
Return the minimum total cost to supply water to all houses.
Example 1:
![](https://assets.leetcode.com/uploads/2019/05/22/1359_ex1.png)
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.
Example 2:
Input: n = 2, wells = [1,1], pipes = [[1,2,1],[1,2,2]] Output: 2 Explanation: We can supply water with cost two using one of the three options: Option 1: - Build a well inside house 1 with cost 1. - Build a well inside house 2 with cost 1. The total cost will be 2. Option 2: - Build a well inside house 1 with cost 1. - Connect house 2 with house 1 with cost 1. The total cost will be 2. Option 3: - Build a well inside house 2 with cost 1. - Connect house 1 with house 2 with cost 1. The total cost will be 2. Note that we can connect houses 1 and 2 with cost 1 or with cost 2 but we will always choose the cheapest option.
Constraints:
2 <= n <= 104
wells.length == n
0 <= wells[i] <= 105
1 <= pipes.length <= 104
pipes[j].length == 3
1 <= house1j, house2j <= n
0 <= costj <= 105
house1j != house2j
Solution(javascript)
/**
* @param {number} n
* @param {number[]} wells
* @param {number[][]} pipes
* @return {number}
*/
var minCostToSupplyWater = function(n, wells, pipes) {
for(let i = 0; i < wells.length; i++) {
pipes.push([0, i + 1, wells[i]])
}
pipes.sort((a, b) => a[2] - b[2])
let res = 0, p = []
for(let [u, v, cost] of pipes) {
let pu = find(u), pv = find(v)
if(pu !== pv) {
p[pu] = pv
res += cost
if(--n === 0) return res
}
}
function find(x) {
if(p[x] === undefined) p[x] = x
while(x !== p[x]) {
p[x] = p[p[x]]
x = p[x]
}
return x
}
};