图 最小生成树 leetcode 1584. 连接所有点的最小费用

https://leetcode.cn/problems/min-cost-to-connect-all-points/

Prim

class Solution {

public:
    // prime 
    int minCostConnectPoints(vector<vector<int>>& points) {
        int res = 0;
        int n = points.size();
        using Pair = pair<int,int>; //(-weight, node v);
        vector<vector<Pair>> g(n);
        vector<int> seen(n, 0);

        if(n <= 1) return 0;

        // 1. construct graph
        for(int i = 0; i < n; i ++){
            for(int j = i + 1; j < n; j ++){
                int w = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
                g[i].push_back({w, j});
                g[j].emplace_back(w, i);
            }
        }

        //2. get minimal edge by priority queue
        priority_queue<Pair> p;
        p.push({0, 0});
        while(n > 0) {
            const int w = -p.top().first;
            const int v = p.top().second;
            p.pop();

            if(seen[v]++) continue;
            for(auto& u: g[v]){
                if(seen[u.second]) continue;
                p.emplace(-u.first, u.second);
            }

            n--;
            res += w; // weight;
        }
        return res;
    }
    
};
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Kruskal

class Solution {

public:
    struct Edge{
        int a, b, w;
        bool operator<(const Edge& e) const {
            return w < e.w;
        }
    };

    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        if(n <= 1) return 0;
        //connections每一行是两个顶点和权重
        vector<Edge> connections;
        for(int i = 0; i < n; i ++){
            for(int j = i + 1; j < n; j ++){
                connections.push_back({i, j, 
                abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1])});
            }
        }
        return minimumCost(n, connections);
    }
    //Kruskal 
    int minimumCost(int n, vector<Edge>& connections) {
        vector<int> parent = vector<int>(n);
        iota(begin(parent), end(parent),0);
        //把所有边按权值进行排序
        sort(connections.begin(), connections.end());
        function<int(int)> find = [&](int x) {
            return x == parent[x] ? x : parent[x] = find(parent[parent[x]]);
        };
        int res = 0;
        for (Edge& conn : connections) {
            int pa = find(conn.a);
            int pb = find(conn.b);
            if (pa == pb) continue;
            parent[pa] = pb;
            res += conn.w;
            if(--n==1) return res;
        }
        return -1;
    }
};
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