本文共 3037 字，大约阅读时间需要 10 分钟。

为了解决这个问题，我们需要构建一个最小生成树，其中必须有一条边连接特定的Nike店和Apple店。我们将使用Prim算法来实现这一点，因为Prim算法能够有效地处理边权重不同的情况，并且适用于较小的点数范围。

方法思路

解决代码

import mathdef main():    import sys    input = sys.stdin.read().split()    ptr = 0    N = int(input[ptr])    ptr += 1    while N != 0:        p = int(input[ptr])        q = int(input[ptr+1])        ptr += 2        pos = [[0.0 for _ in range(2)] for __ in range(N+1)]        for i in range(1, N+1):            pos[i][0] = float(input[ptr])            pos[i][1] = float(input[ptr+1])            ptr += 2        # Initialize distance matrix        a = [[math.inf for _ in range(N+1)] for __ in range(N+1)]        for i in range(1, N+1):            for j in range(i, N+1):                if i == j:                    a[i][j] = 0.0                else:                    a[i][j] = math.hypot(pos[i][0] - pos[j][0], pos[i][1] - pos[j][1])        a[p][q] = 0.0  #强制连接p和q        a[q][p] = a[p][q]        # Prim算法实现        def lct(n, sta):            lowcost = 0.0            # 初始化closeedge数组            closeedge = [ {'vis': False, 'dis': math.inf} for _ in range(n+1)]            closeedge[sta]['vis'] = True            closeedge[p]['vis'] = True            closeedge[q]['vis'] = True            for j in range(1, n+1):                if j != sta and j not in [p, q]:                    closeedge[j]['dis'] = a[sta][j]            # 计算初始lowcost            lowcost += a[p][q]            # 更新其他节点的最短距离            for j in range(1, n+1):                if j not in [p, q] and closeedge[j]['vis'] == False:                    if a[q][j] < closeedge[j]['dis']:                        closeedge[j]['dis'] = a[q][j]            # 开始循环找最短边            for i in range(3, n+1):                min_dist = math.inf                min_j = -1                for j in range(1, n+1):                    if closeedge[j]['vis'] == False and closeedge[j]['dis'] < min_dist:                        min_dist = closeedge[j]['dis']                        min_j = j                if min_j == -1:                    break                lowcost += min_dist                closeedge[min_j]['vis'] = True                # 更新其他节点的最短距离                for j in range(1, n+1):                    if closeedge[j]['vis'] == False and a[min_j][j] < closeedge[j]['dis']:                        closeedge[j]['dis'] = a[min_j][j]            return lowcost        total_cost = lct(N, p)        # 输出结果，保留两位小数        print("{0:.2f}".format(total_cost))        N = int(input[ptr])    returnif __name__ == "__main__":    main()

代码解释

通过这种方法，我们可以确保在构建最小生成树时，特定的两点（Nike店和Apple店）之间有一条边，并且总权重最小。

转载地址：http://mdkzz.baihongyu.com/