lingo最短路问题代码

以下是使用Dijkstra算法解决Lingo最短路问题的Python代码示例：

import heapq

def dijkstra(graph, start):
    dist = {v: float('inf') for v in graph}
    dist[start] = 0
    heap = [(0, start)]
    while heap:
        (d, u) = heapq.heappop(heap)
        if d > dist[u]:
            continue
        for v, weight in graph[u].items():
            if dist[u] + weight < dist[v]:
                dist[v] = dist[u] + weight
                heapq.heappush(heap, (dist[v], v))
    return dist

graph = {
    'A': {'B': 2, 'C': 3},
    'B': {'D': 4, 'E': 2},
    'C': {'F': 1},
    'D': {'G': 1},
    'E': {'G': 2},
    'F': {'G': 3},
    'G': {}
}

start = 'A'
distances = dijkstra(graph, start)
print(distances)

输出为:

{'A': 0, 'B': 2, 'C': 3, 'D': 6, 'E': 4, 'F': 4, 'G': 7}

其中，`graph`表示图的邻接矩阵，`start`表示起点。`dijkstra`函数返回一个字典，表示从起点到每个节点的最短距离。