生成python代码,要求:以邻接矩阵的形式,定义一个简单路网,定义一个起点和一个终点,运用dijkstra算法生成最短路径,然后将最短路径上的所有路段距离增加2倍,然后更新路网,寻找最短路径,直至找到10条最短路径，筛选掉重复的路径，得到不包含重复路径的路径集，以下形式输出路径集：n×m的矩阵，n为路径数量，m为路段数量，如果路段m在路径n里面，则(n,m)值为1，否则为0

以下是生成python代码的答案，可以满足您的要求：

import heapq

# 定义邻接矩阵表示路网
network = [[0, 2, 0, 4, 0, 0],
           [2, 0, 3, 0, 0, 0],
           [0, 3, 0, 1, 3, 0],
           [4, 0, 1, 0, 1, 2],
           [0, 0, 3, 1, 0, 1],
           [0, 0, 0, 2, 1, 0]]

# 定义起点和终点
start, end = 0, 5

# Dijkstra算法求最短路径
def dijkstra(network, start, end):
    n = len(network)
    visited = [False] * n
    distance = [float('inf')] * n
    distance[start] = 0
    heap = [(0, start)]
    while heap:
        (dist, u) = heapq.heappop(heap)
        if visited[u]:
            continue
        visited[u] = True
        for v in range(n):
            if network[u][v] > 0:
                alt = dist + network[u][v]
                if alt < distance[v]:
                    distance[v] = alt
                    heapq.heappush(heap, (alt, v))
    return distance[end]

# 将最短路径上的所有路段距离增加2倍，并更新路网
def double_distance(network, path):
    for i in range(len(path) - 1):
        u, v = path[i], path[i+1]
        network[u][v] *= 2
        network[v][u] *= 2

# 寻找10条最短路径，并筛选掉重复的路径
paths = []
for i in range(10):
    # Dijkstra算法求最短路径
    distance = dijkstra(network, start, end)
    path = [end]
    while path[-1] != start:
        for u in range(len(network)):
            if network[u][path[-1]] > 0 and distance - network[u][path[-1]] == distance[u]:
                path.append(u)
                break
    path.reverse()
    paths.append(path)
    double_distance(network, path)

unique_paths = []
for path in paths:
    if path not in unique_paths:
        unique_paths.append(path)

# 输出路径集
n = len(unique_paths)
m = len(network)
matrix = [[0] * m for i in range(n)]
for i in range(n):
    for j in range(m):
        if j < m - 1 and unique_paths[i][j] == unique_paths[i][j+1]:
            continue
        matrix[i][j] = 1
print(matrix)

希望这份代码能够满足您的需求！