如何实现从起点坐标开始遍历到八个不同坐标的藏宝点，最后再到出口坐标，并且要用时最短给出python代码现已经知道这十个坐标的位置，也有两个坐标之间的最短路径的函数

这是一个典型的最短路径问题，可以使用Dijkstra算法或A*算法来解决。

以A*算法为例，可以先将起点作为起始节点，目标点作为终止节点，对于每个节点，计算其到起点的距离和到目标点的估计距离（启发函数），然后按照总距离从小到大排序，依次遍历相邻节点，如果发现更短的路径，则更新距离和路径，并重新排序。

具体实现过程如下：

import heapq

def heuristic(a, b):
    # 启发函数，估计从a到b的距离
    return abs(a[0] - b[0]) + abs(a[1] - b[1])

def astar(graph, start, end):
    # A*算法实现
    frontier = []  # 待探索节点
    heapq.heappush(frontier, (0, start))  # 将起点加入待探索队列
    came_from = {}  # 记录路径
    cost_so_far = {start: 0}  # 记录起点到每个节点的距离

    while frontier:
        current = heapq.heappop(frontier)[1]  # 取出距离起点最近的节点
        if current == end:
            break

        for next in graph.neighbors(current):
            new_cost = cost_so_far[current] + graph.cost(current, next)
            if next not in cost_so_far or new_cost < cost_so_far[next]:
                cost_so_far[next] = new_cost
                priority = new_cost + heuristic(end, next)  # 计算总距离
                heapq.heappush(frontier, (priority, next))
                came_from[next] = current

    path = [end]
    while path[-1] != start:
        path.append(came_from[path[-1]])
    path.reverse()

    return path

class Graph:
    # 图的定义
    def __init__(self, nodes):
        self.nodes = nodes

    def neighbors(self, node):
        # 返回相邻节点
        return node.adjacent_nodes

    def cost(self, a, b):
        # 返回a到b的距离
        return a.distance_to(b)

class Node:
    # 节点的定义
    def __init__(self, x, y):
        self.x = x
        self.y = y
        self.adjacent_nodes = []
        self.distances = {}

    def add_neighbor(self, node):
        # 添加相邻节点和距离
        self.adjacent_nodes.append(node)
        self.distances[node] = heuristic(self, node)

    def distance_to(self, node):
        # 返回与另一个节点的距离
        return self.distances[node]

# 初始化节点
start_node = Node(0, 0)
treasure1_node = Node(1, 1)
treasure2_node = Node(2, 2)
treasure3_node = Node(3, 3)
treasure4_node = Node(4, 4)
treasure5_node = Node(5, 5)
treasure6_node = Node(6, 6)
treasure7_node = Node(7, 7)
treasure8_node = Node(8, 8)
end_node = Node(9, 9)

# 添加相邻节点和距离
start_node.add_neighbor(treasure1_node)
start_node.add_neighbor(treasure2_node)
start_node.add_neighbor(treasure3_node)
treasure1_node.add_neighbor(treasure2_node)
treasure1_node.add_neighbor(treasure4_node)
treasure2_node.add_neighbor(treasure3_node)
treasure2_node.add_neighbor(treasure4_node)
treasure2_node.add_neighbor(treasure5_node)
treasure3_node.add_neighbor(treasure5_node)
treasure4_node.add_neighbor(treasure6_node)
treasure5_node.add_neighbor(treasure7_node)
treasure6_node.add_neighbor(treasure7_node)
treasure7_node.add_neighbor(treasure8_node)
treasure8_node.add_neighbor(end_node)

# 构建图
graph = Graph([start_node, treasure1_node, treasure2_node, treasure3_node,
               treasure4_node, treasure5_node, treasure6_node, treasure7_node,
               treasure8_node, end_node])

# 计算路径
path1 = astar(graph, start_node, treasure1_node)
path2 = astar(graph, treasure1_node, treasure2_node)
path3 = astar(graph, treasure2_node, treasure3_node)
path4 = astar(graph, treasure1_node, treasure4_node)
path5 = astar(graph, treasure2_node, treasure4_node)
path6 = astar(graph, treasure2_node, treasure5_node)
path7 = astar(graph, treasure3_node, treasure5_node)
path8 = astar(graph, treasure4_node, treasure6_node)
path9 = astar(graph, treasure5_node, treasure7_node)
path10 = astar(graph, treasure7_node, treasure8_node)
path11 = astar(graph, treasure8_node, end_node)

# 输出路径
print("Path from start to treasure1:", path1)
print("Path from treasure1 to treasure2:", path2)
print("Path from treasure2 to treasure3:", path3)
print("Path from treasure1 to treasure4:", path4)
print("Path from treasure2 to treasure4:", path5)
print("Path from treasure2 to treasure5:", path6)
print("Path from treasure3 to treasure5:", path7)
print("Path from treasure4 to treasure6:", path8)
print("Path from treasure5 to treasure7:", path9)
print("Path from treasure7 to treasure8:", path10)
print("Path from treasure8 to end:", path11)

注意，这里的节点距离是使用启发函数估计的，可能不是实际距离，但可以保证结果是最短路径。同时，这里只是求出了每个节点到相邻节点的最短路径，并没有考虑整个路径的连通性，需要在实际应用中根据情况进行调整。