diff options
Diffstat (limited to 'shortest_path_vis_nodes.cpp')
-rw-r--r-- | shortest_path_vis_nodes.cpp | 62 |
1 files changed, 0 insertions, 62 deletions
diff --git a/shortest_path_vis_nodes.cpp b/shortest_path_vis_nodes.cpp deleted file mode 100644 index 8ac5587..0000000 --- a/shortest_path_vis_nodes.cpp +++ /dev/null @@ -1,62 +0,0 @@ -// 847. Shortest Path Visiting All Nodes -#include "leetcode.h" - -/* - * you have an undirected, connected graph of 'n' nodes labeled from 0 to 'n - - * 1'. you are given an array 'graph' where 'graph[i]' is a list of all the - * nodes connected with node 'i' by an edge. return the length of the shortest - * path that visits every node. you may start and stop any any node. you may - * revisit nodes multiple times and you may reused edges. - */ - -class Solution { -public: - class tuple { - public: - int node; - int mask; - int cost; - tuple(int node, int mask, int cost) { - this->node = node; - this->mask = mask; - this->cost = cost; - } - }; - - int shortestPathLength(vvd(int) & graph) { - int n = graph.size(); - queue<tuple> q; - set<pair<int, int>> vis; - int all = (1 << n) - 1; - for (int i = 0; i < n; i++) { - int mask_val = 1 << i; - tuple curr(i, mask_val, 1); - q.push(curr); - vis.insert({i, mask_val}); - } - while (!q.empty()) { - tuple curr = q.front(); - q.pop(); - if (curr.mask == all) - return curr.cost - 1; - for (auto adj : graph[curr.node]) { - int both_vis = curr.mask; - both_vis = both_vis | (1 << adj); - tuple node(adj, both_vis, curr.cost + 1); - if (vis.find({adj, both_vis}) == vis.end()) { - vis.insert({adj, both_vis}); - q.push(node); - } - } - } - return -1; - } -}; - -int main() { - Solution obj; - vvd(int) g1 = {{1, 2, 3}, {0}, {0}}, - g2 = {{1}, {0, 2, 4}, {1, 3, 4}, {2}, {1, 2}}; - printf("%d\n", obj.shortestPathLength(g1)); // expect: 4 - printf("%d\n", obj.shortestPathLength(g2)); // expect: 4 -} |