{ "data": { "question": { "questionId": "1493", "questionFrontendId": "1377", "categoryTitle": "Algorithms", "boundTopicId": 132560, "title": "Frog Position After T Seconds", "titleSlug": "frog-position-after-t-seconds", "content": "

Given an undirected tree consisting of n vertices numbered from 1 to n. A frog starts jumping from vertex 1. In one second, the frog jumps from its current vertex to another unvisited vertex if they are directly connected. The frog can not jump back to a visited vertex. In case the frog can jump to several vertices, it jumps randomly to one of them with the same probability. Otherwise, when the frog can not jump to any unvisited vertex, it jumps forever on the same vertex.

\n\n

The edges of the undirected tree are given in the array edges, where edges[i] = [ai, bi] means that exists an edge connecting the vertices ai and bi.

\n\n

Return the probability that after t seconds the frog is on the vertex target. Answers within 10-5 of the actual answer will be accepted.

\n\n

 

\n

Example 1:

\n\"\"\n
\nInput: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4\nOutput: 0.16666666666666666 \nExplanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 probability to the vertex 2 after second 1 and then jumping with 1/2 probability to vertex 4 after second 2. Thus the probability for the frog is on the vertex 4 after 2 seconds is 1/3 * 1/2 = 1/6 = 0.16666666666666666. \n
\n\n

Example 2:

\n\"\"\n\n
\nInput: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 1, target = 7\nOutput: 0.3333333333333333\nExplanation: The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 = 0.3333333333333333 probability to the vertex 7 after second 1. \n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "T 秒后青蛙的位置", "translatedContent": "

给你一棵由 n 个顶点组成的无向树,顶点编号从 1 到 n。青蛙从 顶点 1 开始起跳。规则如下:

\n\n\n\n

无向树的边用数组 edges 描述,其中 edges[i] = [fromi, toi] 意味着存在一条直接连通 fromitoi 两个顶点的边。

\n\n

返回青蛙在 t 秒后位于目标顶点 target 上的概率。

\n\n

 

\n\n

示例 1:

\n\n

\n\n
\n输入:n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4\n输出:0.16666666666666666 \n解释:上图显示了青蛙的跳跃路径。青蛙从顶点 1 起跳,第 1 秒 有 1/3 的概率跳到顶点 2 ,然后第 2 秒 有 1/2 的概率跳到顶点 4,因此青蛙在 2 秒后位于顶点 4 的概率是 1/3 * 1/2 = 1/6 = 0.16666666666666666 。 \n
\n\n

示例 2:

\n\n

\n\n
\n输入:n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 1, target = 7\n输出:0.3333333333333333\n解释:上图显示了青蛙的跳跃路径。青蛙从顶点 1 起跳,有 1/3 = 0.3333333333333333 的概率能够 1 秒 后跳到顶点 7 。 \n
\n\n

 

\n\n

 

\n\n

提示:

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(listof (listof exact-integer?)) exact-integer? exact-integer? flonum?)\n\n )", "__typename": "CodeSnippetNode" }, { "lang": "Erlang", "langSlug": "erlang", "code": "-spec frog_position(N :: integer(), Edges :: [[integer()]], T :: integer(), Target :: integer()) -> float().\nfrog_position(N, Edges, T, Target) ->\n .", "__typename": "CodeSnippetNode" }, { "lang": "Elixir", "langSlug": "elixir", "code": "defmodule Solution do\n @spec frog_position(n :: integer, edges :: [[integer]], t :: integer, target :: integer) :: float\n def frog_position(n, edges, t, target) do\n\n end\nend", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"4.8K\", \"totalSubmission\": \"14.5K\", \"totalAcceptedRaw\": 4798, \"totalSubmissionRaw\": 14540, \"acRate\": \"33.0%\"}", "hints": [ "Use a variation of DFS with parameters 'curent_vertex' and 'current_time'.", "Update the probability considering to jump to one of the children vertices." ], "solution": null, "status": null, "sampleTestCase": "7\n[[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]]\n2\n4", "metaData": "{\n \"name\": \"frogPosition\",\n \"params\": [\n {\n \"name\": \"n\",\n \"type\": \"integer\"\n },\n {\n \"type\": \"integer[][]\",\n \"name\": \"edges\"\n },\n {\n \"type\": \"integer\",\n \"name\": \"t\"\n },\n {\n \"type\": \"integer\",\n \"name\": \"target\"\n }\n ],\n \"return\": {\n \"type\": \"double\"\n }\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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