📅  最后修改于: 2023-12-03 15:28:41.920000             🧑  作者: Mango
考虑一个电路,其中每个门都有一个开放时间和一个关闭时间。如果在一个门的开放时间内到达的信号的值改变了,那么这个值会被转发到下一个门上。在一个门关闭之后,它不再转发信号。一个门的转发时间是0。给定一组开放时间和关闭时间,在这个电路中计算信号从第一个门到最后一个门所需的最短时间。
以以下格式输入电路信息:
n(电路中的门数)
o1 c1 (第1个门的开放时间和关闭时间)
.
.
on cn (第n个门的开放时间和关闭时间)
其中,n
是一个正整数,表示电路中的门数。o1 c1
到 on cn
是每个门的开放时间和关闭时间。所有时间值都是非负整数值。
输入:
3
1 2
2 3
3 3
输出:
5
本问题是一个经典的最短路问题。可以用Dijkstra算法求解。对于每个门,都可以看做是一个节点,并且相邻的节点有一条边权为开放时间的边连接。可以通过维护一个优先队列来进行最短路的计算。具体步骤如下:
import heapq
def dijkstra(graph, start, end):
# 初始化距离为正无穷
distances = {vertex: float('inf') for vertex in graph}
# 起始门距离为0
distances[start] = 0
# 初始化优先队列
pq = [(0, start)]
while pq:
# 取出距离最小的门
current_distance, current_vertex = heapq.heappop(pq)
# 跳过已经走过的点
if current_distance > distances[current_vertex]:
continue
# 更新与该门相邻的门的距离
for neighbor, weight in graph[current_vertex].items():
distance = current_distance + weight
if distance < distances[neighbor]:
distances[neighbor] = distance
heapq.heappush(pq, (distance, neighbor))
# 目标门的距离被更新
if current_vertex == end:
return distances[end]
# 队列为空,无法到达目标门
return float('inf')
if __name__ == '__main__':
n = int(input())
gates = {}
for i in range(1, n+1):
o, c = map(int, input().split())
gates[i] = {'open_time': o, 'close_time': c}
# 构建门之间的连边
graph = {}
for i in range(1, n+1):
graph[i] = {}
for j in range(i+1, n+1):
if gates[i]['close_time'] + abs(i-j) + gates[j]['open_time'] <= gates[j]['close_time']:
graph[i][j] = abs(i-j)
# 计算起始门(1)到目标门(n)的最短路
print(dijkstra(graph, 1, n))
```python
import heapq
def dijkstra(graph, start, end):
# 初始化距离为正无穷
distances = {vertex: float('inf') for vertex in graph}
# 起始门距离为0
distances[start] = 0
# 初始化优先队列
pq = [(0, start)]
while pq:
# 取出距离最小的门
current_distance, current_vertex = heapq.heappop(pq)
# 跳过已经走过的点
if current_distance > distances[current_vertex]:
continue
# 更新与该门相邻的门的距离
for neighbor, weight in graph[current_vertex].items():
distance = current_distance + weight
if distance < distances[neighbor]:
distances[neighbor] = distance
heapq.heappush(pq, (distance, neighbor))
# 目标门的距离被更新
if current_vertex == end:
return distances[end]
# 队列为空,无法到达目标门
return float('inf')
if __name__ == '__main__':
n = int(input())
gates = {}
for i in range(1, n+1):
o, c = map(int, input().split())
gates[i] = {'open_time': o, 'close_time': c}
# 构建门之间的连边
graph = {}
for i in range(1, n+1):
graph[i] = {}
for j in range(i+1, n+1):
if gates[i]['close_time'] + abs(i-j) + gates[j]['open_time'] <= gates[j]['close_time']:
graph[i][j] = abs(i-j)
# 计算起始门(1)到目标门(n)的最短路
print(dijkstra(graph, 1, n))