73 lines
1.9 KiB
C++
73 lines
1.9 KiB
C++
#include <algorithm>
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#include <vector>
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#include <queue>
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#include <unordered_set>
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class DisjointSet {
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private:
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std::vector<int> parent;
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public:
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explicit DisjointSet(int n) {
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parent.reserve(n);
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for (int i = 0; i < n; ++i)
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parent.push_back(i);
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}
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int Find(int x) {
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return (parent[x] == x) ? x : (parent[x] = Find(parent[x]));
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}
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void Union(int x, int y) {
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auto xR = Find(x), yR = Find(y);
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if (xR != yR)
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parent[xR] = yR;
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}
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};
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struct TupleCmp {
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bool operator()(const std::tuple<int, int, int>& x, const std::tuple<int, int, int>& y) const {
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return std::get<2>(x) > std::get<2>(y);
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}
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};
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/**
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* 1584. Min Cost to Connect All Points
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* You are given an array points representing integer coordinates of some points on a 2D-plane, where points[i] = [xi, yi].
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* 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.
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* Return the minimum cost to make all points connected. All points are connected if there is exactly one simple path between any two points.
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*/
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class Solution {
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public:
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static int minCostConnectPoints(std::vector<std::vector<int>>& points) {
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int n = points.size();
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if (n <= 1)
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return 0;
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std::vector<std::tuple<int, int, int>> edgePairs;
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for (int i = 0; i < n; ++i)
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for (int j = i + 1; j < n; ++j)
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edgePairs.emplace_back(i, j, std::abs(points[i][0] - points[j][0]) + std::abs(points[i][1] - points[j][1]));
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std::priority_queue<std::tuple<int, int, int>, std::vector<std::tuple<int, int, int>>, TupleCmp> q(edgePairs.begin(), edgePairs.end());
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DisjointSet s(n);
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int ans = 0, cnt = 0;
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while (cnt < n - 1) {
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auto cur = q.top();
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q.pop();
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auto i = std::get<0>(cur);
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auto j = std::get<1>(cur);
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if (s.Find(i) != s.Find(j)) {
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s.Union(i, j);
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ans += std::get<2>(cur);
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++cnt;
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}
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}
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return ans;
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}
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};
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