Floyd-Warshall算法:获得最短路径

问题描述

假设图形由 n x n 维邻接矩阵表示.我知道如何获取所有对的最短路径矩阵.但是我不知道有没有办法追踪所有最短的路径?Blow是python代码的实现.

Assume a graph is represented by a n x n dimension adjacency matrix. I know the how to get the shortest path matrix for all pairs. But I wonder is there a way to trace all the shortest paths?Blow is the python code implementation.

v = len(graph)
for k in range(0,v):
    for i in range(0,v):
        for j in range(0,v):
            if graph[i,j] > graph[i,k] + graph[k,j]:
                graph[i,j] = graph[i,k] + graph[k,j]

推荐答案

您必须在if语句中添加一个新的矩阵来存储路径重建数据(数组 p 是前一个矩阵):

You have to add to your if statement a new matrix to store path reconstruction data (array p which is predecessor matrix):

    if graph[i,j] > graph[i,k] + graph[k,j]:
        graph[i,j] = graph[i,k] + graph[k,j]
        p[i,j] = p[k,j]

在开始时，矩阵 p 必须填写为:

At the beginning the matrix p have to be filled as:

for i in range(0,v):
    for j in range(0,v):
        p[i,j] = i
        if (i != j and graph[i,j] == 0):
             p[i,j] = -30000  # any big negative number to show no arc (F-W does not work with negative weights)

要重建 i 和 j 节点之间的路径，您必须调用:

To reconstruct the path between i and j nodes you have to call:

def ConstructPath(p, i, j):
    i,j = int(i), int(j)
    if(i==j):
      print (i,)
    elif(p[i,j] == -30000):
      print (i,'-',j)
    else:
      ConstructPath(p, i, p[i,j]);
      print(j,)

以及具有以上功能的测试:

And the test with above function:

import numpy as np

graph = np.array([[0,10,20,30,0,0],[0,0,0,0,0,7],[0,0,0,0,0,5],[0,0,0,0,10,0],[2,0,0,0,0,4],[0,5,7,0,6,0]])

v = len(graph)

# path reconstruction matrix
p = np.zeros(graph.shape)
for i in range(0,v):
    for j in range(0,v):
        p[i,j] = i
        if (i != j and graph[i,j] == 0):
            p[i,j] = -30000
            graph[i,j] = 30000 # set zeros to any large number which is bigger then the longest way

for k in range(0,v):
    for i in range(0,v):
        for j in range(0,v):
            if graph[i,j] > graph[i,k] + graph[k,j]:
                graph[i,j] = graph[i,k] + graph[k,j]
                p[i,j] = p[k,j]

# show p matrix
print(p)

# reconstruct the path from 0 to 4
ConstructPath(p,0,4)

输出:

p:

[[ 0.  0.  0.  0.  5.  1.]
 [ 4.  1.  5.  0.  5.  1.]
 [ 4.  5.  2.  0.  5.  2.]
 [ 4.  5.  5.  3.  3.  4.]
 [ 4.  5.  5.  0.  4.  4.]
 [ 4.  5.  5.  0.  5.  5.]]

路径0-4:

0
1
5
4

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