生成python代码，要求：定义一个简单路网，定义一个起点和一个终点，运用dijkstra算法得到最短路径，然后将这条路径上的所有路段距离增大2倍，更新路网，继续运用dijkstra算法寻找最短路径，直至寻找到10条路径，将这10条路径里重复的路径删除，生成并输出路径集

以下是符合要求的Python代码：

import heapq

# 定义简单路网
graph = {'A': {'B': 5, 'C': 1},
         'B': {'A': 5, 'C': 2, 'D': 1},
         'C': {'A': 1, 'B': 2, 'D': 4},
         'D': {'B': 1, 'C': 4, 'E': 1},
         'E': {'D': 1}}

# 定义起点和终点
start, end = 'A', 'E'

# 运用dijkstra算法得到最短路径
def dijkstra(graph, start, end):
    # 初始化距离字典和前驱节点字典
    dist = {node: float('inf') for node in graph}
    dist[start] = 0
    prev = {node: None for node in graph}

    # 使用堆实现优先队列
    pq = [(0, start)]
    while pq:
        # 取出距离最小的节点
        (min_dist, curr) = heapq.heappop(pq)
        # 如果当前节点已经处理过，跳过
        if min_dist > dist[curr]:
            continue
        # 遍历当前节点的邻居节点
        for neighbor, weight in graph[curr].items():
            # 计算到邻居节点的距离
            distance = dist[curr] + weight
            # 如果计算出来的距离更小，更新距离和前驱节点
            if distance < dist[neighbor]:
                dist[neighbor] = distance
                prev[neighbor] = curr
                heapq.heappush(pq, (distance, neighbor))
    # 返回最短路径
    path = []
    curr = end
    while curr:
        path.append(curr)
        curr = prev[curr]
    path.reverse()
    return path

# 将最短路径上的所有路段距离增大2倍，更新路网
def update_graph(graph, path):
    for i in range(len(path)-1):
        node1, node2 = path[i], path[i+1]
        graph[node1][node2] *= 2
        graph[node2][node1] *= 2

# 寻找10条最短路径
paths = []
for i in range(10):
    path = dijkstra(graph, start, end)
    paths.append(path)
    # 将最短路径上的所有路段距离增大2倍，更新路网
    update_graph(graph, path)

# 删除重复路径
unique_paths = []
for path in paths:
    if path not in unique_paths:
        unique_paths.append(path)

# 输出路径集
for i, path in enumerate(unique_paths):
    print(f"Path {i+1}: {'->'.join(path)}")