API/api.medcify.app/node_modules/dijkstrajs/dijkstra.js

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2022-09-26 06:11:44 +00:00
'use strict';
/******************************************************************************
* Created 2008-08-19.
*
* Dijkstra path-finding functions. Adapted from the Dijkstar Python project.
*
* Copyright (C) 2008
* Wyatt Baldwin <self@wyattbaldwin.com>
* All rights reserved
*
* Licensed under the MIT license.
*
* http://www.opensource.org/licenses/mit-license.php
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
* THE SOFTWARE.
*****************************************************************************/
var dijkstra = {
single_source_shortest_paths: function(graph, s, d) {
// Predecessor map for each node that has been encountered.
// node ID => predecessor node ID
var predecessors = {};
// Costs of shortest paths from s to all nodes encountered.
// node ID => cost
var costs = {};
costs[s] = 0;
// Costs of shortest paths from s to all nodes encountered; differs from
// `costs` in that it provides easy access to the node that currently has
// the known shortest path from s.
// XXX: Do we actually need both `costs` and `open`?
var open = dijkstra.PriorityQueue.make();
open.push(s, 0);
var closest,
u, v,
cost_of_s_to_u,
adjacent_nodes,
cost_of_e,
cost_of_s_to_u_plus_cost_of_e,
cost_of_s_to_v,
first_visit;
while (!open.empty()) {
// In the nodes remaining in graph that have a known cost from s,
// find the node, u, that currently has the shortest path from s.
closest = open.pop();
u = closest.value;
cost_of_s_to_u = closest.cost;
// Get nodes adjacent to u...
adjacent_nodes = graph[u] || {};
// ...and explore the edges that connect u to those nodes, updating
// the cost of the shortest paths to any or all of those nodes as
// necessary. v is the node across the current edge from u.
for (v in adjacent_nodes) {
if (adjacent_nodes.hasOwnProperty(v)) {
// Get the cost of the edge running from u to v.
cost_of_e = adjacent_nodes[v];
// Cost of s to u plus the cost of u to v across e--this is *a*
// cost from s to v that may or may not be less than the current
// known cost to v.
cost_of_s_to_u_plus_cost_of_e = cost_of_s_to_u + cost_of_e;
// If we haven't visited v yet OR if the current known cost from s to
// v is greater than the new cost we just found (cost of s to u plus
// cost of u to v across e), update v's cost in the cost list and
// update v's predecessor in the predecessor list (it's now u).
cost_of_s_to_v = costs[v];
first_visit = (typeof costs[v] === 'undefined');
if (first_visit || cost_of_s_to_v > cost_of_s_to_u_plus_cost_of_e) {
costs[v] = cost_of_s_to_u_plus_cost_of_e;
open.push(v, cost_of_s_to_u_plus_cost_of_e);
predecessors[v] = u;
}
}
}
}
if (typeof d !== 'undefined' && typeof costs[d] === 'undefined') {
var msg = ['Could not find a path from ', s, ' to ', d, '.'].join('');
throw new Error(msg);
}
return predecessors;
},
extract_shortest_path_from_predecessor_list: function(predecessors, d) {
var nodes = [];
var u = d;
var predecessor;
while (u) {
nodes.push(u);
predecessor = predecessors[u];
u = predecessors[u];
}
nodes.reverse();
return nodes;
},
find_path: function(graph, s, d) {
var predecessors = dijkstra.single_source_shortest_paths(graph, s, d);
return dijkstra.extract_shortest_path_from_predecessor_list(
predecessors, d);
},
/**
* A very naive priority queue implementation.
*/
PriorityQueue: {
make: function (opts) {
var T = dijkstra.PriorityQueue,
t = {},
key;
opts = opts || {};
for (key in T) {
if (T.hasOwnProperty(key)) {
t[key] = T[key];
}
}
t.queue = [];
t.sorter = opts.sorter || T.default_sorter;
return t;
},
default_sorter: function (a, b) {
return a.cost - b.cost;
},
/**
* Add a new item to the queue and ensure the highest priority element
* is at the front of the queue.
*/
push: function (value, cost) {
var item = {value: value, cost: cost};
this.queue.push(item);
this.queue.sort(this.sorter);
},
/**
* Return the highest priority element in the queue.
*/
pop: function () {
return this.queue.shift();
},
empty: function () {
return this.queue.length === 0;
}
}
};
// node.js module exports
if (typeof module !== 'undefined') {
module.exports = dijkstra;
}