以下代码来自https://www.geeksforgeeks.org/shortest-path-exactly-k-edges-directed-weighted-graph/.所有功劳归于PranchalK.
我正在处理生成k边最短路径的问题.以下代码给出了带有预定义k的最短“距离”.
但是,我需要“路径”
对于下面的代码,路径似乎是:
0-> 2-> 3.
编辑:Ajay的代码解决了此问题.但是,每个节点仅需要访问一次.我没有在原始问题中提到这一点.我包括一个额外的数据集来对其进行测试.
# Python3 program to find shortest path
# with exactly k edges
# Define number of vertices in the graph
# and inifinite value
# A naive recursive function to count
# walks from u to v with k edges
def shortestPath(graph,u,v,k):
V = 4
INF = 999999999999
# Base cases
if k == 0 and u == v:
return 0
if k == 1 and graph[u][v] != INF:
return graph[u][v]
if k <= 0:
return INF
# Initialize result
res = INF
# Go to all adjacents of u and recur
for i in range(V):
if graph[u][i] != INF and u != i and v != i:
rec_res = shortestPath(graph,i,k - 1)
if rec_res != INF:
res = min(res,graph[u][i] + rec_res)
return res
# Driver Code
if __name__ == '__main__':
INF = 999999999999
# Let us create the graph shown
# in above diagram
graph = [[0,10,3,2],[INF,INF,7],6],0]]
u = 0
v = 3
k = 2
print("Weight of the shortest path is",shortestPath(graph,k))
# This code is contributed by PranchalK
预期结果是:
[0,2,3]
0是开始节点,3是结束节点.边的数量是2.路径是0->. 2-> 3
编辑:阿杰的答案非常非常接近.但是,每个节点仅需要访问一次.对不起,我本来没有提到这一点.这是要测试的更大数据.
graph = [[0,4,1],[1,7,[2,8,6,[4,1,[3,[7,3]]
Weight of the shortest path is 14
Shortest path is [0,3]
最佳答案
存储产生最小值的节点.每个边缘长度的重量总和< k.
原文链接:https://www.f2er.com/python/533140.htmldef shortestPath(graph,k):
V = 4
INF = 999999999999
# Base cases
if k == 0 and u == v:
return 0,[]
if k == 1 and graph[u][v] != INF:
return graph[u][v],[]
if k <= 0:
return INF,[]
# Initialize result
res = INF
# Go to all adjacents of u and recur
k_minus_one_path = []
least_sum_node = None
for i in range(V):
if graph[u][i] != INF and u != i and v != i:
rec_res,path = shortestPath(graph,k - 1)
if rec_res != INF:
if res > graph[u][i] + rec_res:
k_minus_one_path = path
least_sum_node = i
res = graph[u][i] + rec_res
if least_sum_node is not None:
k_minus_one_path.insert(0,least_sum_node)
k_path = k_minus_one_path
return res,k_path
# Driver Code
if __name__ == '__main__':
INF = 999999999999
# Let us create the graph shown
# in above diagram
graph = [[0,0]]
u = 0
v = 3
k = 2
weight,k)
if weight != INF:
path.insert(0,u) # source
path.append(v) # Destination
print("Weight of the shortest path is",weight)
print("Shortest path is ",path)