本次题目全部来自卡码网
94. 城市间货物运输 I（Bellman_ford 队列优化算法）
bellman_ford使用队列优化
python
import collections
class Edge:
def __init__(self, to, val):
self.to = to # 链接的节点
self.val = val # 边的权重
if __name__ == '__main__':
n, m = map(int, input().split())
grid = [[] for _ in range(n + 1)] # 邻接表
for i in range(m):
p1, p2, val = map(int, input().split())
grid[p1].append(Edge(p2, val))
start = 1 # 起点
end = n # 终点
minDist = [float('inf')] * (n + 1)
minDist[start] = 0
que = collections.deque()
que.append(start) # 队列里放入起点
while que:
node = que.popleft()
for edge in grid[node]:
_from = node
to = edge.to
value = edge.val
if minDist[to] > minDist[_from] + value:
minDist[to] = minDist[_from] + value
que.append(to)
if minDist[end] == float('inf'):
print('unconnected')
else:
print(minDist[end])95. 城市间货物运输 II（bellman_ford之判断负权回路）
检测负权回路
之前讲过的两种做法都能
第一种：松弛n-1次之后继续松弛，最短距离有变化，说明有负权回路，代码超时
第二种：加入队列>=n次，说明有负权回路
第一种
python
if __name__ == '__main__':
n, m = map(int, input().split())
grid = []
for i in range(m):
p1, p2, val = map(int, input().split())
grid.append((p1, p2, val))
start = 1 # 起点
end = n # 终点
minDist = [float('inf')] * (n + 1)
minDist[start] = 0
flag = False
for i in range(n): # 松弛n次，最后一次判断距离是否有变化
for side in grid: # 每一次松弛，都是对所有边进行松弛
_from = side[0] # 边的出发点
to = side[1] # 边的到达点
price = side[2] # 边的权值
# 松弛操作
# minDist[from] != INT_MAX 防止从未计算过的节点出发
if i < n - 1:
if minDist[_from] != float('inf') and minDist[to] > minDist[_from] + price:
minDist[to] = minDist[_from] + price
else:
if minDist[_from] != float('inf') and minDist[to] > minDist[_from] + price:
flag = True
if flag:
print('circle')
elif minDist[end] == float('inf'):
print("unconnected") # 不能到达终点
else:
print(minDist[end]) # 到达终点最短路径第二种
python
import collections
class Edge:
def __init__(self, to, val):
self.to = to # 链接的节点
self.val = val # 边的权重
if __name__ == '__main__':
n, m = map(int, input().split())
grid = [[] for _ in range(n + 1)] # 邻接表
for i in range(m):
p1, p2, val = map(int, input().split())
grid[p1].append(Edge(p2, val))
start = 1 # 起点
end = n # 终点
minDist = [float('inf')] * (n + 1)
minDist[start] = 0
count = [0] * (n + 1)
flag = False
que = collections.deque()
que.append(start) # 队列里放入起点
count[start] += 1
while que:
node = que.popleft()
for edge in grid[node]:
_from = node
to = edge.to
value = edge.val
if minDist[to] > minDist[_from] + value:
minDist[to] = minDist[_from] + value
que.append(to)
count[to] += 1
if count[to] == n:
flag = True
print("circle")
exit()
if minDist[end] == float('inf'):
print('unconnected')
else:
print(minDist[end])96. 城市间货物运输 III（bellman_ford之单源有限最短路）
至于经过k个节点
下面代码超时
python
if __name__ == '__main__':
n, m = map(int, input().split())
grid = []
for i in range(m):
p1, p2, val = map(int, input().split())
grid.append((p1, p2, val))
src, dst, k = map(int, input().split())
minDist = [float('inf')] * (n + 1)
minDist[src] = 0
for i in range(k + 1):
minDist_copy = minDist[:] # 获取上一次计算的结果
for side in grid:
_from = side[0]
to = side[1]
price = side[2]
# 注意使用 minDist_copy 来计算 minDist
if minDist_copy[_from] != float('inf') and minDist[to] > minDist_copy[_from] + price:
minDist[to] = minDist_copy[_from] + price
if minDist[dst] == float('inf'):
print("unreachable") # 不能到达终点
else:
print(minDist[dst]) # 到达终点最短路径下面代码可通过
python
import collections
class Edge:
def __init__(self, to, val):
self.to = to # 链接的节点
self.val = val # 边的权重
if __name__ == '__main__':
n, m = map(int, input().split())
grid = [[] for _ in range(n + 1)] # 邻接表
for i in range(m):
p1, p2, val = map(int, input().split())
grid[p1].append(Edge(p2, val))
start, end, k = map(int, input().split())
k += 1
minDist = [float('inf')] * (n + 1)
minDist[start] = 0
flag = False
que = collections.deque()
que.append(start) # 队列里放入起点
while k != 0 and que:
visited = [False] * (n + 1) # 每一轮松弛中，控制节点不用重复入队列
minDist_copy = minDist[:] # 用来记录每一次遍历的结果
que_size = len(que)
for _ in range(que_size):
node = que.popleft()
for edge in grid[node]:
_from = node
to = edge.to
value = edge.val
if minDist[to] > minDist_copy[_from] + value:
minDist[to] = minDist_copy[_from] + value
if visited[to]: # 不用重复放入队列，但需要重复松弛，所以放在这里位置
continue
visited[to] = True
que.append(to)
k -= 1
if minDist[end] == float('inf'):
print('unreachable')
else:
print(minDist[end])