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Diffstat (limited to 'other/min_cost_walk_weight_graph.cpp')
-rw-r--r-- | other/min_cost_walk_weight_graph.cpp | 72 |
1 files changed, 0 insertions, 72 deletions
diff --git a/other/min_cost_walk_weight_graph.cpp b/other/min_cost_walk_weight_graph.cpp deleted file mode 100644 index 79f3a8b..0000000 --- a/other/min_cost_walk_weight_graph.cpp +++ /dev/null @@ -1,72 +0,0 @@ -// 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"); -} |