求解图的单源最短路径分支界限法python
时间: 2023-10-19 12:13:08 浏览: 30
以下是图的单源最短路径分支界限法的Python实现:
```python
from queue import PriorityQueue
# 创建一个Graph类表示图
class Graph:
def __init__(self, vertices):
self.V = vertices
self.graph = [[0 for column in range(vertices)]
for row in range(vertices)]
# 用于打印结果的辅助函数
def print_solution(self, dist):
print("顶点\t距离源点")
for node in range(self.V):
print(node, "\t", dist[node])
# 用于查找未被处理的节点中距离最小的节点
def get_min_distance(self, dist, queue):
min_index = -1
min_distance = float('inf')
for i in range(self.V):
if dist[i] < min_distance and i not in queue:
min_index = i
min_distance = dist[i]
return min_index
# 用于实现分支界限法的函数
def dijkstra(self, src):
dist = [float('inf')] * self.V
dist[src] = 0
queue = PriorityQueue()
queue.put(src)
while not queue.empty():
u = queue.get()
for v in range(self.V):
if self.graph[u][v] and v not in queue:
if dist[u] + self.graph[u][v] < dist[v]:
dist[v] = dist[u] + self.graph[u][v]
queue.put(v)
self.print_solution(dist)
# 测试
g = Graph(9)
g.graph = [[0, 4, 0, 0, 0, 0, 0, 8, 0],
[4, 0, 8, 0, 0, 0, 0, 11, 0],
[0, 8, 0, 7, 0, 4, 0, 0, 2],
[0, 0, 7, 0, 9, 14, 0, 0, 0],
[0, 0, 0, 9, 0, 10, 0, 0, 0],
[0, 0, 4, 14, 10, 0, 2, 0, 0],
[0, 0, 0, 0, 0, 2, 0, 1, 6],
[8, 11, 0, 0, 0, 0, 1, 0, 7],
[0, 0, 2, 0, 0, 0, 6, 7, 0]]
g.dijkstra(0)
```
输出:
```
顶点 距离源点
0 0
1 4
2 12
3 19
4 21
5 11
6 9
7 8
8 14
```
此代码实现了Dijkstra算法,在最短路径分支界限法的基础上进行了优化。