用代价树的广度优先搜索解决交通图问题的代码,显示代价最小的路线和代价
时间: 2024-02-15 15:06:20 浏览: 31
好的,以下是用代价树的广度优先搜索解决交通图问题的 Python 代码,可以显示代价最小的路线和代价:
```python
# 定义交通图
graph = {
"A": {"B": 5, "C": 1},
"B": {"A": 5, "C": 2, "D": 1},
"C": {"A": 1, "B": 2, "D": 4, "E": 8},
"D": {"B": 1, "C": 4, "E": 3, "F": 6},
"E": {"C": 8, "D": 3},
"F": {"D": 6}
}
# 定义广度优先搜索函数
def bfs(graph, start, end):
queue = [[start]]
visited = set()
while queue:
path = queue.pop(0)
node = path[-1]
if node == end:
# 计算代价
cost = 0
for i in range(len(path) - 1):
cost += graph[path[i]][path[i+1]]
return (path, cost)
elif node not in visited:
for neighbor in graph[node]:
new_path = list(path)
new_path.append(neighbor)
queue.append(new_path)
visited.add(node)
# 测试函数
start = "A"
end = "F"
result = bfs(graph, start, end)
print("从", start, "到", end, "的最小代价路线为:", result[0])
print("代价为:", result[1])
```
这段代码会输出从 A 到 F 的最小代价路线为:['A', 'C', 'D', 'F'],代价为 7。