// 1697. Checking Existence of Edge Length Limited Paths #include "leetcode.h" /* * an undirected graph of 'n' nodes is defined by 'edge_list', * where 'edge_list[i] = [ui, vi, disi]' denotes an edge between * nodes 'ui' and 'vi' with distance 'disi'. note that there may * be multiple edges between two nodes. given an array 'queries', * where 'queries[j] = [pj, qj, limitj]', your task is to determine * for each 'queries[j]' whthere there is a path between 'pj' and * 'qj' such that each edge on the path has a distance strictly less * than 'limitj'. return boolean array where 'ans.length == queries.len' * and the j'th value of 'ans' is true if there is a path for * 'queries[j]' is true, and false otherwise */ class UnionFind { vector parent; public: UnionFind(int n) : parent(n) { iota(begin(parent), end(parent), 0); } void connect(int a, int b) { int x = find(a), y = find(b); if (x == y) return; parent[x] = y; } bool connected(int i, int j) { return find(i) == find(j); } int find(int a) { return parent[a] == a ? a : (parent[a] = find(parent[a])); } }; class Solution { public: vector distanceLimitedPathsExist(int n, vvd & edge_list, vvd & queries) { vector ans(queries.size()); for (int i = 0; i < queries.size(); ++i) queries[i].push_back(i); sort(begin(queries), end(queries), [&](auto &a, auto &b) { return a[2] < b[2]; }); sort(begin(edge_list), end(edge_list), [&](auto &a, auto &b) { return a[2] < b[2]; }); UnionFind uf(n); int i = 0; for (auto &q : queries) { int u = q[0], v = q[1], limit = q[2], qid = q[3]; for (; i < edge_list.size() && edge_list[i][2] < limit; ++i) uf.connect(edge_list[i][0], edge_list[i][1]); ans[qid] = uf.connected(u, v); } return ans; } }; int main() { Solution obj; vvd el1 = {{0, 1, 2}, {1, 2, 4}, {2, 0, 8}, {1, 0, 6}}; vvd q1 = {{0, 1, 2}, {0, 2, 5}}; vector dlpe1 = obj.distanceLimitedPathsExist(3, el1, q1); vvd el2 = {{0, 1, 10}, {1, 2, 5}, {2, 3, 9}, {3, 4, 13}}; vvd q2 = {{0, 4, 14}, {1, 4, 13}}; vector dlpe2 = obj.distanceLimitedPathsExist(5, el2, q2); for (int i = 0; i < 2; i++) cout << dlpe1[i] << ' '; // expect: 0 1 printf("\n"); for (int i = 0; i < 2; i++) cout << dlpe2[i] << ' '; // expect: 1 0 }