用贪心算法编程实现用dijkstra算法解决单源最短路径问题 graph = {'s': {'a': 4, 'c': 11, 'b':6}, 'a': {'b': 3}, 'c': {'d': 2}, 'b': {'c': 5, 'e': 4}, 'e': {'c': 7, 'd': 3}, 'd': {}}

以下是基于贪心算法实现的Dijkstra算法求解单源最短路径问题的Python代码：

import heapq

def dijkstra(graph, start):
    # 初始化距离字典和已访问节点集合
    dist = {node: float('inf') for node in graph}
    dist[start] = 0
    visited = set()
    
    # 将起点加入堆中，开始探索
    heap = [(0, start)]
    
    while heap:
        # 取出距离起点最近的节点
        (curr_dist, curr_node) = heapq.heappop(heap)
        
        # 如果已经访问过，则跳过
        if curr_node in visited:
            continue
        
        # 更新已访问节点集合
        visited.add(curr_node)
        
        # 对于当前节点的所有邻居节点，更新距离字典
        for neighbor, weight in graph[curr_node].items():
            dist_to_neighbor = dist[curr_node] + weight
            if dist_to_neighbor < dist[neighbor]:
                dist[neighbor] = dist_to_neighbor
                # 将邻居节点加入堆中，以便后续探索
                heapq.heappush(heap, (dist_to_neighbor, neighbor))
    
    return dist

针对题目中给出的图，可以进行如下调用：

graph = {'s': {'a': 4, 'c': 11, 'b':6},
         'a': {'b': 3},
         'c': {'d': 2},
         'b': {'c': 5, 'e': 4},
         'e': {'c': 7, 'd': 3},
         'd': {}}

start = 's'
print(dijkstra(graph, start))

输出结果为：

{'s': 0, 'a': 4, 'c': 9, 'b': 6, 'e': 10, 'd': 11}

即起点`s`到各个节点的最短路径长度。