指定起点和终点的有向图算法，找到访问最多节点且权重最小的路径。

使用Dijkstra算法，在最短路径的基础上加上访问节点数量的限制。具体实现如下：

import heapq

def dijkstra(graph, start, end, limit):
    dist = {node: float('inf') for node in graph}
    dist[start] = 0
    visited = set()
    pq = [(0, start, [])]
    
    while pq:
        (d, node, path) = heapq.heappop(pq)
        if node in visited:
            continue
        if node == end:
            return (d, path + [node])
        visited.add(node)
        for neighbor, weight in graph[node].items():
            if neighbor in visited:
                continue
            new_dist = dist[node] + weight
            new_path = path + [node]
            if new_dist > dist[neighbor] or len(new_path) > limit:
                continue
            dist[neighbor] = new_dist
            heapq.heappush(pq, (new_dist, neighbor, new_path))
    return float('inf')

其中，graph为有向图，用字典表示；start为起点，end为终点；limit为访问节点数量的限制。

例如，对于下面的有向图：

graph = {
    'A': {'B': 1, 'C': 2},
    'B': {'C': 3, 'D': 2},
    'C': {'D': 1},
    'D': {}
}

从起点A到终点D，要求访问节点数不超过3个，最短路径是A->B->D，访问节点数为2。我们可以调用函数来计算：

d, path = dijkstra(graph, 'A', 'D', 3)
print("最短路径是", path, "，路径长度为", d)

输出结果为：

最短路径是 ['A', 'B',