我有一个方向图,想要找到从节点A到节点B的最短路径.我在 crates.io上搜索,发现 petgraph看起来像是最受欢迎的箱子.它 implements a number of algorithms,但没有一个能解决我的任务.我错过了什么?
例如,Dijkstra’s algorithm返回路径成本,但哪条路径的成本最低? Bellman-Ford algorithm返回路径成本
和节点,但没有路径.
这是我从图中打印路径的最简单方法:
extern crate petgraph;
use petgraph::prelude::*;
use petgraph::algo::dijkstra;
fn main() {
let mut graph = Graph::<&str, i32>::new();
let a = graph.add_node("a");
let b = graph.add_node("b");
let c = graph.add_node("c");
let d = graph.add_node("d");
graph.extend_with_edges(&[(a, b, 1), (b, c, 1), (c, d, 1), (a, b, 1), (b, d, 1)]);
let paths_cost = dijkstra(&graph, a, Some(d), |e| *e.weight());
println!("dijkstra {:?}", paths_cost);
let mut path = vec![d.index()];
let mut cur_node = d;
while cur_node != a {
let m = graph
.edges_directed(cur_node, Direction::Incoming)
.map(|edge| paths_cost.get(&edge.source()).unwrap())
.min()
.unwrap();
let v = *m as usize;
path.push(v);
cur_node = NodeIndex::new(v);
}
for i in path.iter().rev().map(|v| graph[NodeIndex::new(*v)]) {
println!("path: {}", i);
}
}
据我所知,我需要根据结果自己计算路径
Dijkstra算法.
我相信,如果我自己实现dijkstra(基于我在dijkstra.rs的实现),并且与前一个版本取消注释,并返回前一个版本,那么最终版本将更快,因为答案将类似于前任[前任[d]].
最佳答案 作为 mentioned in the comments(由图书馆的主要作者,不少),您可以使用A *(astar)算法:
extern crate petgraph;
use petgraph::prelude::*;
use petgraph::algo;
fn main() {
let mut graph = Graph::new();
let a = graph.add_node("a");
let b = graph.add_node("b");
let c = graph.add_node("c");
let d = graph.add_node("d");
graph.extend_with_edges(&[(a, b, 1), (b, c, 1), (c, d, 1), (a, b, 1), (b, d, 1)]);
let path = algo::astar(
&graph,
a, // start
|n| n == d, // is_goal
|e| *e.weight(), // edge_cost
|_| 0, // estimate_cost
);
match path {
Some((cost, path)) => {
println!("The total cost was {}: {:?}", cost, path);
}
None => println!("There was no path"),
}
}
The total cost was 2: [NodeIndex(0), NodeIndex(1), NodeIndex(3)]