已知一无向连通图G采用邻接矩阵存储写算法，输出Vi到Vj所经结点数最少的路径

你可以使用Dijkstra算法来解决这个问题。下面是一个基于邻接矩阵的Dijkstra算法的示例代码：

INF = float('inf')  # 无穷大

def dijkstra(graph, start, end):
    # 初始化距离数组和路径数组
    distance = [INF] * len(graph)
    distance[start] = 0
    visited = [False] * len(graph)
    path = [-1] * len(graph)

    for _ in range(len(graph)):
        # 找到当前距离最小的节点
        min_dist = INF
        min_dist_idx = -1
        for i in range(len(graph)):
            if not visited[i] and distance[i] < min_dist:
                min_dist = distance[i]
                min_dist_idx = i

        visited[min_dist_idx] = True

        # 更新与最小距离节点相邻节点的距离
        for j in range(len(graph)):
            if not visited[j] and graph[min_dist_idx][j] > 0 and distance[j] > distance[min_dist_idx] + graph[min_dist_idx][j]:
                distance[j] = distance[min_dist_idx] + graph[min_dist_idx][j]
                path[j] = min_dist_idx

    # 构造最短路径
    shortest_path = []
    curr = end
    while curr != -1:
        shortest_path.append(curr)
        curr = path[curr]

    shortest_path.reverse()

    return shortest_path

# 示例图的邻接矩阵表示
graph = [
    [0, 2, 0, 4, 0],
    [2, 0, 1, 0, 0],
    [0, 1, 0, 3, 0],
    [4, 0, 3, 0, 1],
    [0, 0, 0, 1, 0]
]

start = 0  # Vi的索引
end = 3  # Vj的索引
shortest_path = dijkstra(graph, start, end)
min_node_count = len(shortest_path) - 1

print(f"最短路径为：{shortest_path}")
print(f"最少经过的节点数：{min_node_count}")

在这个例子中，我们使用邻接矩阵表示图G。算法会输出从节点Vi到节点Vj的最短路径，并计算出经过的节点数最少。你可以根据实际情况进行修改和调试。