LeetCode #1168. Optimize Water Distribution in a Village

题目描述：

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]

class Solution {
public:
    static bool comp(const vector<int>& a, const vector<int>& b)
    {
        return a[2]<b[2];
    }
    
    // 利用一个虚拟节点(n+1)，只要这个虚拟节点和其他所有点连通即可，从而转化为最小生成树问题
    int minCostToSupplyWater(int n, vector<int>& wells, vector<vector<int>>& pipes) {
        for(int i=0;i<wells.size();i++) pipes.push_back({i+1,n+1,wells[i]});
        sort(pipes.begin(),pipes.end(),comp); // Kruscal算法将边集排序再用贪心加入边
        
        for(int i=1;i<=n+1;i++) parent[i]=i;
        int cost=0;
        for(vector<int> pipe:pipes)
        {
            int x=find(pipe[0]);
            int y=find(pipe[1]);
            if(x!=y)
            {
                cost+=pipe[2];
                parent[x]=y;
            }
        }
        return cost;
    }
    
    int find(int x)
    {
        if(x==parent[x]) return x;
        else return find(parent[x]);
    }

private:
    unordered_map<int,int> parent;
};