问题描述
假设我有一个无向图（可以是循环的或非循环的），其中每个节点都分配有一个整数状态。我想找到以下路径：
- 遍历每个节点，但只遍历一次
- 不需要遍历每一个边缘
- 最大化每个移动状态变化的总和
例如，我有一个循环图 -5-4-5-7-2-（前 5 个和最后 2 个相连）。如果我们从前 5 个开始，到最后 2 个结束，则每次移动的变化总和将为 -1 + 1 + 2 + (-5) = -3。该图可以用邻接矩阵描述如下：
import numpy as np
node_states = [5,4,5,7,2]
# Adjacency matrix
#5 4 5 7 2
am = np.array([[0,1,1],# 5
[1,0],# 4
[0,# 5
[0,# 7
[1,0]])# 2
预期输出是
max_delta_sum_path = [2,7]
其中路径总和最大3 + (-1) + 1 + 2 = 5
有谁知道有没有比较快的算法可以自动找到这条路径？
解决方法
- 用两个有向、有成本的链接替换每个无向链接。例如，状态 5 和 7 节点之间的链接将被成本为 +2 和 -2 的两条链接替换。
- 为每个链接的成本增加价值，使所有链接成本为正
- 找出最昂贵链接的成本并从每个链接成本中减去
- 将每个链接成本乘以 -1
- 应用旅行商算法
因此，以您的示例为例：
0 -> 1 cost -1 converts to 6
0 -> 4 cost -3 converts to 8
1 -> 0 cost 1 converts to 4
1 -> 2 cost 1 converts to 4
2 -> 1 cost -1 converts to 6
2 -> 3 cost 2 converts to 3
3 -> 2 cost -2 converts to 7
3 -> 4 cost -5 converts to 10
4 -> 0 cost 3 converts to 2
4 -> 3 cost 5 converts to 0
我认为这就是您要找的：
import numpy as np
node_states = [5,4,5,7,2]
# Adjacency matrix
#5 4 5 7 2
am = np.array([[0,1,1],# 5
[1,0],# 4
[0,# 5
[0,# 7
[1,0]])# 2
for i in range(len(node_states)):
for j in range(len(node_states)):
if am[i][j] == 1:
am[i][j] = node_states[i] - node_states[j] # go through ever entry in every list,and if it is 1 replace it with the traversal cost
"""
am = [[ 0 1 0 0 3]
[-1 0 -1 0 0]
[ 0 1 0 -2 0]
[ 0 0 2 0 5]
[-3 0 0 -5 0]]
"""
from itertools import permutations
def largest_sum(node_states,am):
largest = None
largest_journey = None
traversal_list = list(permutations(range(len(node_states)),len(node_states))) # store all possible permutations of node_states indexes
for trav in traversal_list: # go through each permuatation
costs = [] # track the cost of each traversal
for i in range(len(trav)):
if i == 0: # there are one less traversals than nodes so we are ignoring the first node
continue
if am[trav[i]][trav[i-1]] == 0: # traversal cannot happen if the traversal has no adjacency
continue
costs.append(am[trav[i]][trav[i-1]]) # use the updated am matrix to get our costs,and store them here
if len(costs) == len(node_states) - 1: # if one less traversal was made than we have nodes,we know all nodes were visited
costs_sum = sum(costs) # sum the costs for our total of travel
if largest is None or largest < costs_sum: # only keep this total if it was bigger than our old total
largest = costs_sum # track the new total
largest_trav = list(map(lambda x: node_states[x],trav)) # change our array of indexes (trav) into an array of node values
return largest_trav # when the looping is done,return our total
out = largest_sum(node_states,am)
print(out)
输出：
[2,7]