LeetCode 1584. 连接所有点的最小费用

LeetCode 1584. 连接所有点的最小费用

Kruskal 更加适合稀疏图

const int N = 5e5 + 10;
struct Edge
{
    
    int a, b, w;
    bool operator < (Edge& e)
    {
    
        return w < e.w;
    }
}e[N];

class Solution {
    
public:
    int p[N];
    int find(int x)
    {
    
        if(x != p[x]) p[x] = find(p[x]);
        return p[x];
    }
    int kruskal(int n)
    {
    
        sort(e, e + n);
        int res = 0, cnt = 0;
        for(int i = 0; i < n; i ++)
        {
    
            int a = e[i].a, b = e[i].b, w = e[i].w;
            int pa = find(a), pb = find(b);
            if(pa != pb)
            {
    
                p[pa] = pb;
                cnt ++;
                res += w;
            }
            if(cnt == n - 1) break;
        }
        return res;
    }
    int minCostConnectPoints(vector<vector<int>>& points) {
    
        int n = points.size();
        for(int i = 0; i <= n; i ++) p[i] = i;
        int idx = 0;
        for(int i = 0; i < n; i ++)
            for(int j = i + 1; j < n; j ++)
            {
    
                int d = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
                e[idx ++ ] = {
    i, j, d};
            }
        return kruskal(idx);
    }
};

Prim 更加适合稠密图

const int N = 1010, INF = 0x3f3f3f3f;
class Solution {
    
public:
    int g[N][N];
    int dist[N];
    bool st[N] = {
    false};
    int prim(int n)
    {
    
        memset(dist, 0x3f, sizeof dist);
        int res = 0;
        for(int i = 0; i < n; i ++)
        {
    
            int t = -1;
            for(int j = 0; j < n; j ++)
                if(!st[j] && (t == -1 || dist[j] < dist[t]))
                    t = j;
            if(i && dist[t] == INF) return INF;
            if(i) res += dist[t];
            st[t] = true;
            for(int i = 0; i < n; i ++)
                dist[i] = min(dist[i], g[t][i]);
        }
        return res;
    }
    int minCostConnectPoints(vector<vector<int>>& points) {
    
        int n = points.size();
        memset(g, 0x3f, sizeof g);
        for(int i = 0; i < n; i ++)
            for(int j = i + 1; j < n; j ++)
            {
    
                int d = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
                g[i][j] = d;
                g[j][i] = d;
            }
        return prim(n);
    }
};