Leetcode Daily Question

Leetcode-cn.com 2021-08-24
🟡 787.cheapest-flights-within-k-stops

🏷️ Tags
#depth_first_search #breadth_first_search #graph #dynamic_programming #shortest_path #heap_priority_queue

Description
有 n 个城市通过一些航班连接。给你一个数组 flights ，其中 flights[i] = [fromi, toi, pricei] ，表示该航班都从城市 fromi 开始，以价格 toi 抵达 pricei。

现在给定所有的城市和航班，以及出发城市 src 和目的地 dst，你的任务是找到出一条最多经过 k 站中转的路线，使得从 src 到 dst 的价格最便宜，并返回该价格。如果不存在这样的路线，则输出 -1。

Example

输入: 
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
输出: 200
解释: 
城市航班图如下


从城市 0 到城市 2 在 1 站中转以内的最便宜价格是 200，如图中红色所示。

142 views17:10

CN CN Solution

Leetcode.com 2021-09-12
🔴 882.reachable-nodes-in-subdivided-graph

🏷️ Tags
#graph #shortest_path #heap_priority_queue

Description
You are given an undirected graph (the "original graph") with n nodes labeled from 0 to n - 1. You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge.

The graph is given as a 2D array of edges where edges[i] = [ui, vi, cnti] indicates that there is an edge between nodes ui and vi in the original graph, and cnti is the total number of new nodes that you will subdivide the edge into. Note that cnti == 0 means you will not subdivide the edge.

To subdivide the edge [ui, vi], replace it with (cnti + 1) new edges and cnti new nodes. The new nodes are x1, x2, ..., xcnti, and the new edges are [ui, x1], [x1, x2], [x2, x3], ..., [xcnti+1, xcnti], [xcnti, vi].

In this new graph, you want to know how many nodes are reachable from the node 0, where a node is reachable if the distance is maxMoves or less.

Given the original graph and maxMoves, return the number of nodes that are reachable from node 0 in the new graph.

Example

Input: edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3
Output: 13
Explanation: The edge subdivisions are shown in the image above.
The nodes that are reachable are highlighted in yellow.

132 views16:50

EN EN Solution

Leetcode-cn.com 2022-01-24
🔴 2045.second-minimum-time-to-reach-destination

🏷️ Tags
#breadth_first_search #graph #shortest_path

Description
城市用一个双向连通图表示，图中有 n 个节点，从 1 到 n 编号（包含 1 和 n）。图中的边用一个二维整数数组 edges 表示，其中每个 edges[i] = [ui, vi] 表示一条节点 ui 和节点 vi 之间的双向连通边。每组节点对由最多一条边连通，顶点不存在连接到自身的边。穿过任意一条边的时间是 time 分钟。

每个节点都有一个交通信号灯，每 change 分钟改变一次，从绿色变成红色，再由红色变成绿色，循环往复。所有信号灯都同时改变。你可以在任何时候进入某个节点，但是只能在节点信号灯是绿色时才能离开。如果信号灯是绿色，你不能在节点等待，必须离开。

第二小的值是严格大于最小值的所有值中最小的值。

例如，[2, 3, 4] 中第二小的值是 3 ，而 [2, 2, 4] 中第二小的值是 4 。

给你 n、edges、time 和 change ，返回从节点 1 到节点 n 需要的第二短时间。

注意：

你可以任意次穿过任意顶点，包括 1 和 n 。
你可以假设在启程时，所有信号灯刚刚变成绿色。

Example

输入：n = 5, edges = [[1,2],[1,3],[1,4],[3,4],[4,5]], time = 3, change = 5
输出：13
解释：
上面的左图展现了给出的城市交通图。
右图中的蓝色路径是最短时间路径。
花费的时间是：
- 从节点 1 开始，总花费时间=0
- 1 -> 4：3 分钟，总花费时间=3
- 4 -> 5：3 分钟，总花费时间=6
因此需要的最小时间是 6 分钟。

右图中的红色路径是第二短时间路径。
- 从节点 1 开始，总花费时间=0
- 1 -> 3：3 分钟，总花费时间=3
- 3 -> 4：3 分钟，总花费时间=6
- 在节点 4 等待 4 分钟，总花费时间=10
- 4 -> 5：3 分钟，总花费时间=13
因此第二短时间是 13 分钟。

166 views16:30

CN CN Solution

882.reachable-nodes-in-subdivided-graph.pdf

220 KB

leetcode.cn 2022-11-26
🔴882.reachable-nodes-in-subdivided-graph

🏷️ Tags
#graph #shortest_path #heap_priority_queue

201 views02:11

787.cheapest-flights-within-k-stops.pdf

108.5 KB

leetcode.com 2023-01-26
🟡787.cheapest-flights-within-k-stops

🏷️ Tags
#dynamic_programming #depth_first_search #breadth_first_search #graph #heap_priority_queue #shortest_path

198 views17:09

leetcode.com 2023-05-20
🟡399.evaluate-division

🏷️ Tags
#depth_first_search #breadth_first_search #union_find #graph #array #shortest_path

You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i]. Each Ai or Bi is a string that represents a single variable. You are also given…

evaluate-division

147 views00:11

leetcode.cn 2023-06-04
🔴2699.modify-graph-edge-weights

🏷️ Tags
#graph #shortest_path #heap_priority_queue

modify-graph-edge-weights

给你一个 n 个节点的无向带权连通图，节点编号为 0 到 n - 1 ，再给你一个整数数组 edges ，其中 edges[i] = [ai, bi, wi] 表示节点 ai 和 bi 之间有一条边权为 wi 的边。部分边的边权为 -1（wi = -1），其他边的边权都为正数（wi > 0）。你需要将所有边权为 -1 的边都修改为范围 [1, 2 * 109] 中的正整数，使得从节点 source 到节点 destination 的最短距离为整数 target 。如果有多种修改方案可以使 source…

110 views16:07

leetcode.cn 2023-06-09
🔴2699.modify-graph-edge-weights

🏷️ Tags
#graph #shortest_path #heap_priority_queue

modify-graph-edge-weights

133 views16:05

leetcode.com 2023-06-28
🟡1514.path-with-maximum-probability

🏷️ Tags
#graph #array #shortest_path #heap_priority_queue

path-with-maximum-probability

You are given an undirected weighted graph of n nodes (0-indexed), represented by an edge list where edges[i] = [a, b] is an undirected edge connecting the nodes a and b with a probability of success of traversing that edge succProb[i]. Given two nodes start and end…

141 views00:02

leetcode.com 2023-11-11
🔴2642.design-graph-with-shortest-path-calculator

🏷️ Tags
#graph #design #shortest_path #heap_priority_queue

design-graph-with-shortest-path-calculator

There is a directed weighted graph that consists of n nodes numbered from 0 to n - 1. The edges of the graph are initially represented by the given array edges where edges[i] = [fromi, toi, edgeCosti] meaning that there is an edge from fromi to toi with the…

433 views00:08

leetcode.cn 2023-11-14
🟡1334.find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance

🏷️ Tags
#graph #dynamic_programming #shortest_path

find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance

有 n 个城市，按从 0 到 n-1 编号。给你一个边数组 edges，其中 edges[i] = [fromi, toi, weighti] 代表 fromi 和 toi 两个城市之间的双向加权边，距离阈值是一个整数 distanceThreshold。返回能通过某些路径到达其他城市数目最少、且路径距离最大为 distanceThreshold 的城市。如果有多个这样的城市，则返回编号最大的城市。注意，连接城市 i 和 j 的路径的距离等于沿该路径的所有边的权重之和。示例 1：输入：n…

👍1

240 views16:04