leetcode 787 航班最少花费

题目：
There are n cities connected by m flights. Each fight starts from city u and arrives at v with a price w.

Now given all the cities and flights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1.

Example 1:
Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
Output: 200
Explanation:
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the picture.
Example 2:
Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 0
Output: 500
Explanation:
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the picture.
Note:

The number of nodes n will be in range [1, 100], with nodes labeled from 0 to n - 1.
The size of flights will be in range [0, n * (n - 1) / 2].
The format of each flight will be (src, dst, price).
The price of each flight will be in the range [1, 10000].
k is in the range of [0, n - 1].
There will not be any duplicated flights or self cycles.
普通的广度优先搜索：
graph=[[None]*n for _ in range(n)]
for row in flights:
graph[row[0]][row[1]]=row[2]
q,step,minCost=[(src,0)],0,-1
while q:
if step-1K:
break
new=[]
for stop,cost in q:
for Next,c in enumerate(graph[stop]):
if c and Nextdst:
minCost=c+cost if minCost==-1 else min(minCost,cost+c)
elif c and (minCost==-1 or c+cost<=minCost):
new.append((Next,c+cost))
q,step=new,step+1
return minCost
使用优先队列的广度优先搜索：
f,heap=collections.defaultdict(dict),[(0,src,K+1)]
for a,b,c in flights:
f[a][b]=c
while heap:
c,i,k=heapq.heappop(heap)
if i==dst:
return c
if k>0:
for j in f[i]:
heapq.heappush(heap,(c+f[i][j],j,k-1))
return -1