代码找出树的A到F的最短路径以及经过的顺序节点
时间: 2023-05-14 08:05:38 浏览: 40
这个问题可以使用 Dijkstra 算法来解决。以下是示例代码:
```python
import heapq
def dijkstra(graph, start, end):
# 初始化距离和前驱节点
dist = {node: float('inf') for node in graph}
dist[start] = 0
prev = {node: None for node in graph}
# 使用堆来实现优先队列
pq = [(0, start)]
while pq:
# 取出距离最小的节点
(d, node) = heapq.heappop(pq)
# 如果已经找到终点,则返回路径
if node == end:
path = []
while prev[node] is not None:
path.append(node)
node = prev[node]
path.append(start)
path.reverse()
return path
# 更新相邻节点的距离和前驱节点
for neighbor in graph[node]:
new_dist = dist[node] + graph[node][neighbor]
if new_dist < dist[neighbor]:
dist[neighbor] = new_dist
prev[neighbor] = node
heapq.heappush(pq, (new_dist, neighbor))
# 如果无法到达终点,则返回空路径
return []
# 构建图
graph = {
'A': {'B': 2, 'C': 4},
'B': {'A': 2, 'C': 1, 'D': 3},
'C': {'A': 4, 'B': 1, 'D': 2, 'E': 3},
'D': {'B': 3, 'C': 2, 'E': 1, 'F': 5},
'E': {'C': 3, 'D': 1, 'F': 4},
'F': {'D': 5, 'E': 4}
}
# 找出最短路径
path = dijkstra(graph, 'A', 'F')
# 输出结果
if path:
print('最短路径为:', ' -> '.join(path))
else:
print('无法到达终点')
```
输出结果为:
```
最短路径为: A -> B -> D -> F
```
这个路径经过的顺序节点为 A、B、D、F。