实现从起点坐标开始遍历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，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，给出python代码

好的，下面是一个简单的实现，使用 Dijkstra 算法来计算最短路径：

import heapq

def shortest_path(start, end, points, distance_fn):
    # 使用 Dijkstra 算法计算最短路径
    pq = [(0, start)]
    visited = set()
    distances = {start: 0}
    while pq:
        (cost, node) = heapq.heappop(pq)
        if node == end:
            break
        if node in visited:
            continue
        visited.add(node)
        for neighbor in points:
            if neighbor == node or neighbor in visited:
                continue
            new_cost = cost + distance_fn(node, neighbor)
            if neighbor not in distances or new_cost < distances[neighbor]:
                distances[neighbor] = new_cost
                heapq.heappush(pq, (new_cost, neighbor))
    return distances[end]

# 10 个点的坐标
points = [(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9)]

# 两个点之间的最短路径函数
def distance_fn(p1, p2):
    return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1])

# 依次遍历八个藏宝点
start = (0, 0)
end = (9, 9)
total_distance = 0
for i in range(1, 9):
    point = points[i]
    distance = shortest_path(start, point, points, distance_fn)
    total_distance += distance
    start = point
# 最后再到出口坐标
total_distance += shortest_path(start, end, points, distance_fn)

print("最短路径的长度为：", total_distance)

需要注意的是，这里的 `points` 列表中，第一个元素是起点，最后一个元素是终点，中间的 8 个元素是藏宝点。