分支限界法求图的最短回路源码
时间: 2024-03-25 09:39:12 浏览: 74
以下是一个使用分支限界法求解图的最短回路问题的Python源码示例:
```python
import sys
# 定义一个节点类,用于表示状态空间中的节点
class Node:
def __init__(self, path, bound, length):
self.path = path # 当前路径
self.bound = bound # 当前路径的下界
self.length = length # 当前路径已经走过的长度
# 定义一个函数,用于计算两点之间的距离
def distance(x1, y1, x2, y2):
return ((x1-x2)**2 + (y1-y2)**2)**0.5
# 定义一个函数,用于计算当前路径的下界
def lower_bound(graph, path):
n = len(graph)
visited = [False] * n
# 计算已经走过的长度
length = sum([distance(path[i][0], path[i][1], path[i+1][0], path[i+1][1]) for i in range(len(path)-1)])
# 计算剩余路径的最小长度
for i in range(len(path)):
visited[path[i]] = True
bound = length
for i in range(n):
if not visited[i]:
min_dist = sys.maxsize
for j in range(n):
if visited[j]:
dist = graph[i][j]
if dist < min_dist:
min_dist = dist
bound += min_dist
return bound
# 定义一个函数,用于求解图的最短回路问题
def tsp(graph):
n = len(graph)
# 初始化起始节点
start_node = Node([0], lower_bound(graph, [0]), 0)
# 初始化最优解
best_path = None
best_length = sys.maxsize
# 定义一个优先队列,用于存放待扩展的节点
queue = [start_node]
while queue:
# 从队列中取出一个节点
curr_node = queue.pop(0)
# 如果当前节点的下界比当前最优解还要大,剪枝
if curr_node.bound >= best_length:
continue
# 如果当前节点的路径已经包含了所有的节点,更新最优解
if len(curr_node.path) == n:
curr_length = curr_node.length + graph[curr_node.path[-1]][0]
if curr_length < best_length:
best_path = curr_node.path + [0]
best_length = curr_length
# 否则,扩展当前节点的子节点
else:
for i in range(n):
if i not in curr_node.path:
new_path = curr_node.path + [i]
new_bound = lower_bound(graph, new_path)
new_length = curr_node.length + graph[curr_node.path[-1]][i]
new_node = Node(new_path, new_bound, new_length)
# 将子节点插入队列中,并按照下界从小到大的顺序排序
queue.append(new_node)
queue.sort(key=lambda x: x.bound)
return best_path, best_length
# 测试代码
graph = [
[0, 1, 2, 3],
[1, 0, 4, 5],
[2, 4, 0, 6],
[3, 5, 6, 0]
]
best_path, best_length = tsp(graph)
print("best path:", best_path)
print("best length:", best_length)
```
在上述代码中,我们定义了一个`Node`类来表示状态空间中的节点,每个节点包含了当前已经走过的路径、当前路径的下界和已经走过的路径长度。我们使用贪心算法来计算当前路径的下界,然后使用优先队列来存放待扩展的节点,每次从队列中取出下界最小的节点进行扩展。在扩展节点时,如果当前节点的下界比当前最优解还要大,我们就将其剪枝,不再扩展它的子节点。如果当前节点的路径已经包含了所有的节点,我们就更新最优解。最终,我们返回最优解的路径和长度。
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