在Python中实现Dijkstra算法

26

我正在尝试使用数组在Python中实现Dijkstra算法。这是我的实现。

def extract(Q, w):
    m=0
    minimum=w[0]
    for i in range(len(w)):
        if w[i]<minimum:
            minimum=w[i]
            m=i
    return m, Q[m]

def dijkstra(G, s, t='B'):
    Q=[s]
    p={s:None}
    w=[0]
    d={}
    for i in G:
        d[i]=float('inf')
        Q.append(i)
        w.append(d[i])
    d[s]=0
    S=[]
    n=len(Q)
    while Q:
        u=extract(Q,w)[1]
        S.append(u)
        #w.remove(extract(Q, d, w)[0])
        Q.remove(u)
        for v in G[u]:
            if d[v]>=d[u]+G[u][v]:
                d[v]=d[u]+G[u][v]
                p[v]=u
    return d, p

B='B'
A='A'
D='D'
G='G'
E='E'
C='C'
F='F'
G={B:{A:5, D:1, G:2}, A:{B:5, D:3, E:12, F:5}, D:{B:1, G:1, E:1, A:3}, G:{B:2, D:1, C:2}, C:{G:2, E:1, F:16}, E:{A:12, D:1, C:1, F:2}, F:{A:5, E:2, C:16}}
print "Assuming the start vertex to be B:"
print "Shortest distances", dijkstra(G, B)[0]
print "Parents", dijkstra(G, B)[1]

我期望你的答案是:

Assuming the start vertex to be B:
Shortest distances {'A': 4, 'C': 4, 'B': 0, 'E': 2, 'D': 1, 'G': 2, 'F': 4}
Parents {'A': 'D', 'C': 'G', 'B': None, 'E': 'D', 'D': 'B', 'G': 'D', 'F': 'E'}

然而,我得到的答案是:
Assuming the start vertex to be B:
Shortest distances {'A': 4, 'C': 4, 'B': 0, 'E': 2, 'D': 1, 'G': 2, 'F': 10}
Parents {'A': 'D', 'C': 'G', 'B': None, 'E': 'D', 'D': 'B', 'G': 'D', 'F': 'A'}.

对于节点 F,程序给出了错误的答案。请问有人能告诉我原因吗?


21
请使用有意义的变量名称,这将帮助人们更好地理解你的代码。 - shaktimaan
3
你的代码非常混乱:有两个不同名字的变量G,还有一个没用到的变量S等等。我猜你的代码只能找到不超过两条边的路径,因为你从未向队列中添加任何东西(在Dijkstra算法中应该这样做),但由于代码难以阅读,我无法确定。 - Natalya Ginzburg
13个回答

0
这是我使用最小优先队列实现Dijkstra算法的代码,希望对你有用。
from collections import defaultdict
from math import floor


class MinPQ:
    """
    each heap element is in form (key value, object handle), while heap
    operations works based on comparing key value and object handle points to
    the corresponding application object.
    """

    def __init__(self, array=[]):
        self._minheap = list(array)
        self._length = len(array)
        self._heapsize = 0
        self._build_min_heap()

    def _left(self, idx):
        return 2*idx+1

    def _right(self, idx):
        return 2*idx+2

    def _parent(self, idx):
        return floor((idx-1)/2)

    def _min_heapify(self, idx):
        left = self._left(idx)
        right = self._right(idx)
        min_idx = idx
        if left <= self._heapsize-1 and self._minheap[left] < self._minheap[min_idx]:
            min_idx = left
        if right <= self._heapsize-1 and self._minheap[right] < self._minheap[min_idx]:
            min_idx = right
        if min_idx != idx:
            self._minheap[idx], self._minheap[min_idx] = self._minheap[min_idx], self._minheap[idx]
            self._min_heapify(min_idx)

    def _build_min_heap(self):
        self._heapsize = self._length
        mid_id = int(self._heapsize-1)-1
        for i in range(mid_id, -1, -1):
            self._min_heapify(i)

    def decrease_key(self, idx, new_key):
        while idx > 0 and new_key < self._minheap[self._parent(idx)]:
            self._minheap[idx] = self._minheap[self._parent(idx)]
            idx = self._parent(idx)
        self._minheap[idx] = new_key

    def extract_min(self):
        minimum = self._minheap[0]
        self._minheap[0] = self._minheap[self._heapsize-1]
        self._heapsize = self._heapsize - 1
        self._min_heapify(0)
        return minimum

    def insert(self, item):
        self._minheap.append(item)
        self._heapsize = self._heapsize + 1
        self.decrease_key(self._heapsize-1, item)

    @property
    def minimum(self):
        return self._minheap[0]

    def is_empty(self):
        return self._heapsize == 0

    def __str__(self):
        return str(self._minheap)

    __repr__ = __str__

    def __len__(self):
        return self._heapsize


class DiGraph:
    def __init__(self, edges=None):
        self.adj_list = defaultdict(list)
        self.add_weighted_edges(edges)

    @property
    def nodes(self):
        nodes = set()
        nodes.update(self.adj_list.keys())
        for node in self.adj_list.keys():
            for neighbor, weight in self.adj_list[node]:
                nodes.add(neighbor)
        return list(nodes)

    def add_weighted_edges(self, edges):
        if edges is None:
            return None
        for edge in edges:
            self.add_weighted_edge(edge)

    def add_weighted_edge(self, edge):
        node1, node2, weight = edge
        self.adj_list[node1].append((node2, weight))

    def weight(self, tail, head):
        for node, weight in self.adj_list[tail]:
            if node == head:
                return weight
        return None


def relax(min_heapq, dist, graph, u, v):
    if dist[v] > dist[u] + graph.weight(u, v):
        dist[v] = dist[u] + graph.weight(u, v)
        min_heapq.insert((dist[v], v))


def dijkstra(graph, start):
    # initialize
    dist = dict.fromkeys(graph.nodes, float('inf'))
    dist[start] = 0
    min_heapq = MinPQ()
    min_heapq.insert((0, start))

    while not min_heapq.is_empty():
        distance, u = min_heapq.extract_min()
        # we may add a node multiple time in priority queue, but we process it
        # only once
        if distance > dist[u]:
            continue
        for neighbor, weight in graph.adj_list[u]:
            relax(min_heapq, dist, graph, u, neighbor)

    return dist

if __name__ == "__main__":
    edges = [('s', 't', 10), ('t', 'x', 1), ('s', 'y', 5), ('y', 't', 3), ('t', 'y', 2), 
             ('y', 'x', 9), ('y', 'z', 2), ('z', 's', 7), ('x', 'z', 4), ('z', 'x', 6)]
    digraph = DiGraph(edges)
    res = dijkstra(digraph, 's')
    print(res)


0

我使用优先队列实现了Dijkstra算法。除此之外,我还自己实现了最小堆。希望这对你有所帮助。

from collections import defaultdict


class MinPQ:
    """
    each heap element is in form (key value, object handle), while heap
    operations works based on comparing key value and object handle points to
    the corresponding application object.
    """

    def __init__(self, array=[]):
        self._minheap = list(array)
        self._length = len(array)
        self._heapsize = 0
        self._build_min_heap()

    def _left(self, idx):
        return 2*idx+1

    def _right(self, idx):
        return 2*idx+2

    def _parent(self, idx):
        return int((idx-1)/2)

    def _min_heapify(self, idx):
        left = self._left(idx)
        right = self._right(idx)
        min_idx = idx
        if left <= self._heapsize-1 and self._minheap[left] < self._minheap[min_idx]:
            min_idx = left
        if right <= self._heapsize-1 and self._minheap[right] < self._minheap[min_idx]:
            min_idx = right
        if min_idx != idx:
            self._minheap[idx], self._minheap[min_idx] = self._minheap[min_idx], self._minheap[idx]
            self._min_heapify(min_idx)

    def _build_min_heap(self):
        self._heapsize = self._length
        mid_id = int((self._heapsize)/2)-1
        for i in range(mid_id, -1, -1):
            self._min_heapify(i)

    def decrease_key(self, idx, new_key):
        while idx > 0 and new_key < self._minheap[self._parent(idx)]:
            self._minheap[idx] = self._minheap[self._parent(idx)]
            idx = self._parent(idx)
        self._minheap[idx] = new_key

    def extract_min(self):
        if self._heapsize < 1:
            raise IndexError
        minimum = self._minheap[0]
        self._minheap[0] = self._minheap[self._heapsize-1]
        self._heapsize = self._heapsize - 1
        self._min_heapify(0)
        return minimum

    def insert(self, item):
        self._minheap.append(item)
        self._heapsize = self._heapsize + 1
        self.decrease_key(self._heapsize-1, item)

    @property
    def minimum(self):
        return self._minheap[0]

    def is_empty(self):
        return self._heapsize == 0

    def __str__(self):
        return str(self._minheap)

    __repr__ = __str__

    def __len__(self):
        return self._heapsize


class DiGraph:
    def __init__(self, edges=None):
        self.adj_list = defaultdict(list)
        self.add_weighted_edges(edges)

    @property
    def nodes(self):
        nodes = set()
        nodes.update(self.adj_list.keys())
        for node in self.adj_list.keys():
            for neighbor, weight in self.adj_list[node]:
                nodes.add(neighbor)
        return list(nodes)

    def add_weighted_edges(self, edges):
        if edges is None:
            return None
        for edge in edges:
            self.add_weighted_edge(edge)

    def add_weighted_edge(self, edge):
        node1, node2, weight = edge
        self.adj_list[node1].append((node2, weight))

    def weight(self, tail, head):
        for node, weight in self.adj_list[tail]:
            if node == head:
                return weight
        return None


def relax(min_heapq, dist, graph, u, v):
    if dist[v] > dist[u] + graph.weight(u, v):
        dist[v] = dist[u] + graph.weight(u, v)
        min_heapq.insert((dist[v], v))


def dijkstra(graph, start):
    # initialize
    dist = dict.fromkeys(graph.nodes, float('inf'))
    dist[start] = 0
    min_heapq = MinPQ()
    min_heapq.insert((0, start))

    while not min_heapq.is_empty():
        distance, u = min_heapq.extract_min()
        # we may add a node multiple time in priority queue, but we process it
        # only once
        if distance > dist[u]:
            continue
        for neighbor, weight in graph.adj_list[u]:
            relax(min_heapq, dist, graph, u, neighbor)

    return dist

0

在extract函数中设置一个断点。你会发现你从Q中删除了条目,但是从w中却没有删除。除此之外,其他都是字典,但是Q/w是一对数组,你没有及时更新它们。你必须保持这两个同步,或者用字典替换它们。特别注意:最终,在算法工作后,你可能想用边的列表替换Q/w,并使用优先队列(heapq模块)重新编写“extract”函数。

此外,你会发现w始终具有源节点的权重为0,而所有其他节点的权重为'inf'。你完全跳过了更新候选距离的关键步骤。

因此,你基本上总是选择遇到的第一条路径,而不是选择最短路径。你稍后计算该路径的实际距离,因此返回的距离数组具有实际值,但是它们是任意选择的,你没有理由期望它们是最短的。

在你(错误地)找到下一个节点之后,你查看它的所有边。这应该是我在第二段中提到的关键步骤,在那里你更新下一个节点的候选项。相反,你做了完全不同的事情:你似乎正在遍历所有先前的解决方案(如果你正确地实现迪杰斯特拉算法,则保证正确并且应该被保留),并且你正在寻找从源->当前->任何位置的两步解决方案。查看它们的正确意图应该是将来自先前路径的下一个候选项添加到下一个节点,但由于它们从未被添加,因此你只查看字面上的两个节点解决方案,而不是(先前最短路径)+(一步),。
所以基本上,你正在循环遍历从源到所有可能的两个节点路径,以找到最短路径。这是一个完全错误的方法,与迪杰斯特拉算法无关。但是,在你的微小图表中,大多数正确的最短路径都是两步路径,因此它几乎可以工作。
(附言:我同意大家对你的变量名称的看法。如果你使用详细的名称来说明这些变量代表什么,你会做得更好。在分析你的代码之前,我必须将它们重命名。)

网页内容由stack overflow 提供, 点击上面的
可以查看英文原文,
原文链接