leetcode 1377. Frog Position After T Seconds

Given an undirected tree consisting of n vertices numbered from 1 to n. A frog starts jumping from the 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. 

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

Return the probability that after t seconds the frog is on the vertex target.

 

Example 1:

Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4
Output: 0.16666666666666666 
Explanation: 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. 

Example 2:

Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 1, target = 7
Output: 0.3333333333333333
Explanation: 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. 

Example 3:

Input: n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 20, target = 6
Output: 0.16666666666666666

 

Constraints:

• 1 <= n <= 100
• edges.length == n-1
• edges[i].length == 2
• 1 <= edges[i][0], edges[i][1] <= n
• 1 <= t <= 50
• 1 <= target <= n
• Answers within 10^-5 of the actual value will be accepted as correct.

题目 求青蛙从节点1在时刻t跳到节点target的概率。

实际上是一个搜索问题。概率计算是简单的。只需要注意考虑时间不够和跳过头的情况即可。这里使用的是广度优先。代码如下：

class Solution(object):
    def frogPosition(self, n, edges, t, target):
        """
        :type n: int
        :type edges: List[List[int]]
        :type t: int
        :type target: int
        :rtype: float
        """
        if n == 1: 
            return 1.0
        link = [[] for i in xrange(n + 1)]
        link[1].append(1) # 确保通用匹配
        for i, j in edges:
            link[i].append(j)
            link[j].append(i)
        prob = [0] * (n + 1)
        l = [1]
        prob[1] = 1 #第一个位置的概率是1
        step = 0 #轮次，如果时间超过轮次，target又有后续节点。概率为0
        while l:
            next_l = []
            for i in l:
                if i == target:
                    if t < step:# 时间不够
                        return 0
                    if t > step and len(link[target]) > 1:#跳过了
                        return 0
                    return 1.0/prob[target]
                il = len(link[i])-1 # 计算概率是去掉父亲节点
                for j in link[i]:
                    if prob[j] == 0: #还没有访问过
                        prob[j] = prob[i]*il
                        next_l.append(j)
            l = next_l
            step += 1
        return 0