1. 若不考虑时间，可以采用最短路径算法（如Dijkstra算法）来求解最短路径。首先需要建立各考场之间的距离矩阵，然后求出从家到所有考场的最短路径，再求出所有考场之间的最短路径，最后将所有最短路径相加即可得到考完所有场次回到家后的路径最短的路线。如何用代码实现

首先需要定义一个表示图的类，包含节点和边：

class Graph:
    def __init__(self, vertices):
        self.V = vertices
        self.graph = [[0 for column in range(vertices)]
                      for row in range(vertices)]

    def printSolution(self, dist):
        print("Vertex \t Distance from Source")
        for node in range(self.V):
            print(node, "\t\t", dist[node])

    def minDistance(self, dist, sptSet):
        min = float('inf')
        for v in range(self.V):
            if dist[v] < min and sptSet[v] == False:
                min = dist[v]
                min_index = v

        return min_index

    def dijkstra(self, src):
        dist = [float('inf')] * self.V
        dist[src] = 0
        sptSet = [False] * self.V

        for cout in range(self.V):
            u = self.minDistance(dist, sptSet)
            sptSet[u] = True
            for v in range(self.V):
                if self.graph[u][v] > 0 and sptSet[v] == False and \
                        dist[v] > dist[u] + self.graph[u][v]:
                    dist[v] = dist[u] + self.graph[u][v]

        return dist

然后，定义一个函数来计算从家到所有考场的最短路径：

def calculate_shortest_path(home, exam_centers):
    graph = Graph(len(exam_centers) + 1)
    for i, center in enumerate(exam_centers):
        for j, other_center in enumerate(exam_centers):
            if i != j:
                distance = calculate_distance(center, other_center)
                graph.graph[i][j] = distance
                graph.graph[j][i] = distance

    for i, center in enumerate(exam_centers):
        distance = calculate_distance(home, center)
        graph.graph[-1][i] = distance
        graph.graph[i][-1] = distance

    return graph.dijkstra(len(exam_centers))

其中，`calculate_distance` 是一个计算两点之间距离的函数，可以使用经纬度计算两点之间的距离，也可以使用其他的方法。最后，将所有最短路径相加即可得到考完所有场次回到家后的路径最短的路线。