【LeetCode】787. Cheapest Flights Within K Stops 解题报告(Python)
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址:https://leetcode.com/problems/cheapest-flights-within-k-stops/description/
题目描述
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.
题目大意
有N个城市,m个航班,他们之间的连接是个有向图。现在已知最多可以中转k次,求从srt到dst的最小花费。
解题方法
图的遍历的基础上加上了一个限制条件:最多中转k次,即最多只能访问k+1个节点。可以用DFS和BFS两者方法去解决。
方法一:DFS
这个其实就是回溯法,先从起点开始向后搜索,如果搜索到了dst或者没有步数了,那么换下一条路进行搜索。需要使用一个visited数组表示已经搜索过的节点,这样可以防止走成一个环。
另外这个题需要一个强剪枝,就是当某条路径的花费大于了我们当前到达dst需要花费的最小值的时候,后面的路径都不需要走了,这个是由于题目给出的路费都是整数,向下走哪怕走到了dst花费也会更高。
时间复杂度是O(N^2),空间复杂度是O(1).打败了6%的提交。
class Solution(object):
def findCheapestPrice(self, n, flights, src, dst, K):
"""
:type n: int
:type flights: List[List[int]]
:type src: int
:type dst: int
:type K: int
:rtype: int
"""
graph = collections.defaultdict(dict)
for u, v, e in flights:
graph[u][v] = e
visited = [0] * n
ans = [float('inf')]
self.dfs(graph, src, dst, K + 1, 0, visited, ans)
return -1 if ans[0] == float('inf') else ans[0]
def dfs(self, graph, src, dst, k, cost, visited, ans):
if src == dst:
ans[0] = cost
return
if k == 0:
return
for v, e in graph[src].items():
if visited[v]: continue
if cost + e > ans[0]: continue
visited[v] = 1
self.dfs(graph, v, dst, k - 1, cost + e, visited, ans)
visited[v] = 0
方法二:BFS
如果给定步数的情况下,一个更直接的方法就是BFS,这样就可以直接判断在指定的k步以内能不能走到dst,不会进行更多的搜索了,因此这个方法要快很多。
BFS是个模板,直接使用一个队列很容易就实现了。这个队列存放的是当我们进行第step次搜索时,搜索到的当前的节点,以及走到当前节点的花费。所以当当前节点走到dst时,更新最小花费。
时间复杂度是O(KN),空间复杂度是O(N).打败了60%的提交。
class Solution(object):
def findCheapestPrice(self, n, flights, src, dst, K):
"""
:type n: int
:type flights: List[List[int]]
:type src: int
:type dst: int
:type K: int
:rtype: int
"""
graph = collections.defaultdict(dict)
for u, v, e in flights:
graph[u][v] = e
ans = float('inf')
que = collections.deque()
que.append((src, 0))
step = 0
while que:
size = len(que)
for i in range(size):
cur, cost = que.popleft()
if cur == dst:
ans = min(ans, cost)
for v, w in graph[cur].items():
if cost + w > ans:
continue
que.append((v, cost + w))
if step > K: break
step += 1
return -1 if ans == float('inf') else ans
参考资料
https://www.youtube.com/watch?v=PLY-lbcxEjg
日期
2018 年 10 月 23 日 —— 风真是个好东西
【LeetCode】787. Cheapest Flights Within K Stops 解题报告(Python)的更多相关文章
- LeetCode 787. Cheapest Flights Within K Stops
原题链接在这里:https://leetcode.com/problems/cheapest-flights-within-k-stops/ 题目: There are n cities connec ...
- [LeetCode] 787. Cheapest Flights Within K Stops K次转机内的最便宜航班
There are n cities connected by m flights. Each fight starts from city u and arrives at v with a pri ...
- [LeetCode] 787. Cheapest Flights Within K Stops_Medium tag: Dynamic Programming, BFS, Heap
There are n cities connected by m flights. Each fight starts from city u and arrives at v with a pri ...
- 787. Cheapest Flights Within K Stops
There are n cities connected by m flights. Each fight starts from city u and arrives at v with a pri ...
- [LeetCode] Cheapest Flights Within K Stops K次转机内的最便宜的航班
There are n cities connected by m flights. Each fight starts from city u and arrives at v with a pri ...
- [Swift]LeetCode787. K 站中转内最便宜的航班 | Cheapest Flights Within K Stops
There are n cities connected by m flights. Each fight starts from city u and arrives at v with a pri ...
- Within K stops 最短路径 Cheapest Flights Within K Stops
2018-09-19 22:34:28 问题描述: 问题求解: 本题是典型的最短路径的扩展题,可以使用Bellman Ford算法进行求解,需要注意的是在Bellman Ford算法的时候需要额外申请 ...
- 【LeetCode】94. Binary Tree Inorder Traversal 解题报告(Python&C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 解题方法 递归 迭代 日期 题目地址:https://leetcode.c ...
- 【LeetCode】341. Flatten Nested List Iterator 解题报告(Python&C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归+队列 栈 日期 题目地址:https://lee ...
随机推荐
- EXCEL ctrl+e 百变用法不只是你用的那么简单
Excel2013版本中,新增加了一个快捷键:Ctrl+E,可以依据字符之间的关系,实现快速填充功能.一些需要使用公式或者其他功能进行解决的问题,现在只要一个快捷键就可以实现了. 用法1:快速拆解出需 ...
- Perl字符串处理函数用法集锦
Perl字符串处理函数 0.函数名 index 调用语法position=index(string,substring,position); 解说返回子串substring在字符串string中的位置 ...
- Redis | 第9章 Lua 脚本与排序《Redis设计与实现》
目录 前言 1. Lua 脚本 1.1 Redis 创建并修改 Lua 环境的步骤 1.2 Lua 环境协作组件 1.3 EVAL 命令的实现 1.4 EVALSHA 命令的实现 1.5 脚本管理命令 ...
- Levenshtein莱文斯坦算法在项目中的应用
简介 根据维基百科的描述,在信息理论.语言学和计算机科学中,莱文斯坦距离是一个测量两个序列之间差异的字符串度量.非正式地,两个单词之间的莱文斯坦距离是将一个单词改变为另一个单词所需的最小单字符编辑次数 ...
- A Child's History of England.15
And indeed it did. For, the great army landing from the great fleet, near Exeter, went forward, layi ...
- Hadoop【MR开发规范、序列化】
Hadoop[MR开发规范.序列化] 目录 Hadoop[MR开发规范.序列化] 一.MapReduce编程规范 1.Mapper阶段 2.Reducer阶段 3.Driver阶段 二.WordCou ...
- 零基础学习java------day1------计算机基础以及java的一些简单了解
一. java的简单了解 Java是一门面向对象编程语言,不仅吸收了C++的各种优点,还摒弃了C++里难以理解的多继承.指针等概念,因此Java语言具有功能强大和简单易用两个特征.Java语言作为静态 ...
- [PE结构]导入表与IAT表
导入表的结构导入表的结构 typedef struct _IMAGE_IMPORT_DESCRIPTOR { union { DWORD Characteristics; // 0 for termi ...
- 【Linux】【Commands】文本查看类
分屏查看命令:more和less more命令: more FILE 特点:翻屏至文件尾部后自动退出: less命令: less FILE head命令: 查看文件的前n行: head [option ...
- Servlet+Jdbc+mysql实现登陆功能
首先是新建一个servlet,servlet中有dopost和doget方法 一般的表格提交都是用post方法,故在dopost里面写入逻辑代码 下面是其逻辑代码Check.java protecte ...