已知所有点的坐标，只有两点间距离小于a才能连通，python输出所有的连通路径，并绘制所有的连通路径，图上包含每个点的坐标和对应的下标

可以使用深度优先搜索（DFS）来查找所有的连通路径，并使用matplotlib库绘制图形。

以下是实现代码：

import matplotlib.pyplot as plt

def dfs(graph, start, visited, path, a):
    visited.add(start)
    path.append(start)

    for node in graph[start]:
        if node not in visited and distance(start, node) < a:
            dfs(graph, node, visited, path, a)

    if len(path) == len(graph):
        print(path)
        draw_graph(graph, path)

    visited.remove(start)
    path.pop()

def distance(node1, node2):
    x1, y1 = node1
    x2, y2 = node2
    return ((x1 - x2) ** 2 + (y1 - y2) ** 2) ** 0.5

def draw_graph(graph, path):
    plt.figure()
    for i, node in enumerate(graph):
        x, y = node
        plt.text(x, y, str(i) + ":(" + str(x) + "," + str(y) + ")")

    for i in range(len(path) - 1):
        node1 = graph[path[i]]
        node2 = graph[path[i+1]]
        x1, y1 = node1
        x2, y2 = node2
        plt.plot([x1, x2], [y1, y2], 'bo-')

    plt.show()

if __name__ == '__main__':
    # 输入所有点的坐标
    points = [(0,0), (1,1), (2,2), (3,3), (4,4), (5,5)]

    # 计算所有点之间的距离
    graph = {}
    for i, node1 in enumerate(points):
        graph[i] = []
        for j, node2 in enumerate(points):
            if i != j and distance(node1, node2) < 2:
                graph[i].append(j)

    # 查找所有的连通路径，并绘制图形
    visited = set()
    path = []
    dfs(graph, 0, visited, path, 2)

这个程序首先计算所有点之间的距离，然后根据距离构建图。接着，使用深度优先搜索算法查找所有的连通路径，并绘制图形。在绘图时，每个点的坐标和下标都被标注在了图上。