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author | Jack Sangdahl <[email protected]> | 2025-03-30 21:18:48 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2025-03-30 21:18:48 -0600 |
commit | e1b13ac0df29496c79403cbe20c64a3c7eb414b7 (patch) | |
tree | 11f6cd9354b3d30b3d454f1dbf6b8aa26989aaaf /other | |
parent | 62ec80fb2041268d4d7aaad4cbe5b3fc7c67ce3f (diff) | |
download | leetcode-e1b13ac0df29496c79403cbe20c64a3c7eb414b7.tar.gz leetcode-e1b13ac0df29496c79403cbe20c64a3c7eb414b7.zip |
chore: move remaining cpp files to other/
Diffstat (limited to 'other')
-rw-r--r-- | other/min_cost_walk_weight_graph.cpp | 72 | ||||
-rw-r--r-- | other/num_ways_arrive_dest.cpp | 62 |
2 files changed, 134 insertions, 0 deletions
diff --git a/other/min_cost_walk_weight_graph.cpp b/other/min_cost_walk_weight_graph.cpp new file mode 100644 index 0000000..79f3a8b --- /dev/null +++ b/other/min_cost_walk_weight_graph.cpp @@ -0,0 +1,72 @@ +// 3108. Minimum Cost Walk in Weighted Graph +#include "leetcode.h" + +/* + * there is an undirected weighted graph with 'n' vertices labeled from 0 to 'n + * - 1'. you are given the integer 'n' and an array 'edges', where 'edges[i] = + * [u_i, v_i, w_i]' indicates that there is an edge between vertices 'u_i' and + * 'v_i' with a weight of 'w_i'. a walk on the graph is a sequence of vertices + * and edges. the walk starts and ends with a vertex, and each edge connects the + * vertex that comes before it and the vertex that comes after it. it's + * important to note that a walk may visit the same edge or vertex more than + * once. return the array 'answer' where 'answer[i]' denotes the minimum cost of + * a walk for query 'i'. + */ + +class Solution { +public: + vector<int> minimumCost(int n, vector<vector<int>> &edges, + vector<vector<int>> &query) { + vector<vector<int>> adj(n); + vector<int> cnt(n, (1 << 30) - 1); + for (vector<int> e : edges) { + int u = e[0], v = e[1], k = e[2]; + adj[u].push_back(v); + adj[v].push_back(u); + cnt[u] &= k; + cnt[v] &= k; + } + queue<int> q; + vector<int> group(n, -1); + for (int i = 0; i < n; i++) { + if (group[i] == -1) { + group[i] = i; + q.push(i); + while (!q.empty()) { + int u = q.front(); + q.pop(); + cnt[i] &= cnt[u]; + for (int v : adj[u]) { + if (group[v] == -1) { + group[v] = i; + q.push(v); + } + } + } + } + } + vector<int> ans; + for (vector<int> qs : query) { + int s = qs[0], t = qs[1]; + if (s == t) + ans.push_back(0); + else + ans.push_back(group[s] == group[t] ? cnt[group[s]] : -1); + } + return ans; + } +}; + +int main() { + Solution obj; + vvd e1 = {{0, 1, 7}, {1, 3, 7}, {1, 2, 1}}, q1 = {{0, 3}, {3, 4}}; + vvd e2 = {{0, 2, 7}, {0, 1, 15}, {1, 2, 6}, {1, 2, 1}}, q2 = {{1, 2}}; + vector<int> mc1 = obj.minimumCost(5, e1, q1); + vector<int> mc2 = obj.minimumCost(3, e2, q2); + for (auto i : mc1) + printf("%d ", i); // expect: 1 -1 + printf("\n"); + for (auto i : mc2) + printf("%d ", i); // expect: 0 + printf("\n"); +} diff --git a/other/num_ways_arrive_dest.cpp b/other/num_ways_arrive_dest.cpp new file mode 100644 index 0000000..416d811 --- /dev/null +++ b/other/num_ways_arrive_dest.cpp @@ -0,0 +1,62 @@ +// 1976. Number of Ways to Arrive at Destination +#include "leetcode.h" + +/* + * you are in a city that consists of 'n' intersections numbered from 0 to 'n - + * 1' with bi directional roads between some intersections. the inputs are + * generated such that you can reach any intersection from any other + * intersection and that there is at most one road between any two + * intersections. you are given an integer 'n' and a 2d integer array 'roads' + * where 'roads[i] = [u_i, v_i, t_i]' means that there is a road between + * intersections 'u' and 'v' that takes 't' minutes to travel. you want to know + * in how many ways you can travel from interseciton 0 to intersection 'n - 1'. + */ + +class Solution { +public: + int countPaths(int n, vector<vector<int>> &roads) { + vector<vector<pair<long long, long long>>> graph(n); + for (auto &r : roads) { + long long u = r[0], v = r[1], t = r[2]; + graph[u].push_back({v, t}); + graph[v].push_back({u, t}); + } + return dijkstra(graph, n, 0); + } + +private: + int mod = 1e9 + 7; + int dijkstra(const vector<vector<pair<long long, long long>>> &graph, int n, + int src) { + vector<long long> dist(n, LONG_MAX), ways(n); + ways[src] = 1; + dist[src] = 0; + priority_queue<pair<long long, long long>, + vector<pair<long long, long long>>, greater<>> + pq; + pq.push({0, 0}); + while (!pq.empty()) { + auto [d, u] = pq.top(); + pq.pop(); + if (d > dist[u]) + continue; + for (auto [v, t] : graph[u]) { + if (dist[v] > d + t) { + dist[v] = d + t; + ways[v] = ways[u]; + } else if (dist[v] == d + t) + ways[v] = (ways[v] + ways[u]) % mod; + } + } + return ways[n - 1]; + } +}; + +int main() { + Solution obj; + vvd r1 = {{0, 6, 7}, {0, 1, 2}, {1, 2, 3}, {1, 3, 3}, {6, 3, 3}, + {3, 5, 1}, {6, 5, 1}, {2, 5, 1}, {0, 4, 5}, {4, 6, 2}}; + vvd r2 = {{1, 0, 10}}; + printf("%d\n", obj.countPaths(7, r1)); // expect: 4 + printf("%d\n", obj.countPaths(2, r2)); // expect: 1 +} |