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1584. Min Cost to Connect All Points

Description

You are given an array points representing integer coordinates of some points on a 2D-plane, where points[i] = [xi, yi].

The cost of connecting two points [xi, yi] and [xj, yj] is the manhattan distance between them: |xi - xj| + |yi - yj|, where |val| denotes the absolute value of val.

Return the minimum cost to make all points connected. All points are connected if there is exactly one simple path between any two points.

 

Example 1:

Input: points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
Output: 20
Explanation: 

We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.

Example 2:

Input: points = [[3,12],[-2,5],[-4,1]]
Output: 18

 

Constraints:

  • 1 <= points.length <= 1000
  • -106 <= xi, yi <= 106
  • All pairs (xi, yi) are distinct.

Solutions

Solution 1

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class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        n = len(points)
        g = [[0] * n for _ in range(n)]
        dist = [inf] * n
        vis = [False] * n
        for i, (x1, y1) in enumerate(points):
            for j in range(i + 1, n):
                x2, y2 = points[j]
                t = abs(x1 - x2) + abs(y1 - y2)
                g[i][j] = g[j][i] = t
        dist[0] = 0
        ans = 0
        for _ in range(n):
            i = -1
            for j in range(n):
                if not vis[j] and (i == -1 or dist[j] < dist[i]):
                    i = j
            vis[i] = True
            ans += dist[i]
            for j in range(n):
                if not vis[j]:
                    dist[j] = min(dist[j], g[i][j])
        return ans

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class Solution {
    public int minCostConnectPoints(int[][] points) {
        final int inf = 1 << 30;
        int n = points.length;
        int[][] g = new int[n][n];
        for (int i = 0; i < n; ++i) {
            int x1 = points[i][0], y1 = points[i][1];
            for (int j = i + 1; j < n; ++j) {
                int x2 = points[j][0], y2 = points[j][1];
                int t = Math.abs(x1 - x2) + Math.abs(y1 - y2);
                g[i][j] = t;
                g[j][i] = t;
            }
        }
        int[] dist = new int[n];
        boolean[] vis = new boolean[n];
        Arrays.fill(dist, inf);
        dist[0] = 0;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int j = -1;
            for (int k = 0; k < n; ++k) {
                if (!vis[k] && (j == -1 || dist[k] < dist[j])) {
                    j = k;
                }
            }
            vis[j] = true;
            ans += dist[j];
            for (int k = 0; k < n; ++k) {
                if (!vis[k]) {
                    dist[