petgraph 的哪种算法将找到从 A 到 B 的最短路径

Question

>我有一个方向图，想找到从节点A到节点B的最短路径。我在 crates.io 上搜索了一下，发现了看起来像最受欢迎的板条箱的petgraph。它实现了许多算法，但没有一个能解决我的任务。我错过了什么吗？

例如，Dijkstra 的算法返回路径成本，但哪条路径的成本最低？贝尔曼-福特算法返回路径成本和节点，但没有路径。

这是我发现从图形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 .

我相信，如果我自己实现dijkstra(基于 dijkstra.rs 实现(，并取消注释带有predecessor行，并返回predecessor，最终的变体会更快，因为答案将是类似于predecessor[predecessor[d]]。

Answer 1

正如评论中提到的(由图书馆的主要作者，不少(，您可以使用 A* ( astar ( 算法：

use petgraph::{algo, prelude::*}; // 0.5.1
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)]