python第六章课后习题

print("学号：2023310143028")

  
def prim(graph, start):  
    num_nodes = len(graph)  
    visited = [False] * num_nodes  
    min_heap = [(0, start, -1)]   
    mst_cost = 0  
    mst_edges = []  
  
    while min_heap:  
        weight, u, parent = heapq.heappop(min_heap)  
        if visited[u]:  
            continue  
        visited[u] = True  
        mst_cost += weight  
        if parent != -1:  
            mst_edges.append((parent, u, weight))  
  
        for v in range(num_nodes):  
            if not visited[v] and graph[u][v] != 0:  
                heapq.heappush(min_heap, (graph[u][v], v, u))  
  
    return mst_cost, mst_edges  
    
graph = [  
    [0,20,0,0,15,0],  
    [20,0,20,60,25,0],  
    [0,20,0,30,18,0],  
    [0,60,30,0,35,10],  
    [0,0,0,10,15,0]  
]  
  
mst_cost, mst_edges = prim(graph, 0)  
print("Prim's MST Cost:", mst_cost)  
print("Prim's MST Edges:", mst_edges)
 
print("学号：3028")


print("学号：2023310143028")

import heapq  initial_costs = [2.5, 2.6, 2.8, 3.1] 
salvage_values = [2.0, 1.6, 1.3, 1.1]  
maintenance_costs = [0.3, 0.8, 1.5, 2.0]  
 
dp = [[float('inf')] * 2 for _ in range(4)] 
dp[0][1] = initial_costs[0] + maintenance_costs[0]  
  
for i in range(1, 4):  
    dp[i][1] = min(dp[i-1][1] + maintenance_costs[i],  
                   initial_costs[i] + maintenance_costs[i])  
 
    if i > 0:  
        dp[i][0] = dp[i-1][1] + salvage_values[i-1] 
  
min_cost = min(dp[3][1],  
               min(dp[i][0] for i in range(3)))  
  
 
print(f"最优更新策略下的4年内最小总费用为：{min_cost}万元")  
 
print("学号：3028")


print("学号：2023310143028")