https://leetcode.com/problems/path-with-maximum-probability/description/
1514. Path with Maximum Probability
给你一个大小为 n 的无向图资讯，阵列 edges[] 表示边关系，succProb[] 表示边的机
率权重，路径经过时需要乘以该权重，求出从 start 走到 end 的路径，他的最终权重必
需是最高，走不到则返回0。
Example 1:
https://assets.leetcode.com/uploads/2019/09/20/1558_ex1.png
Input: n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start =
0, end = 2
Output: 0.25000
Explanation: There are two paths from start to end, one having a probability
of success = 0.2 and the other has 0.5 * 0.5 = 0.25.
Example 2:
https://assets.leetcode.com/uploads/2019/09/20/1558_ex2.png
Input: n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.3], start =
0, end = 2
Output: 0.30000
Example 3:
https://assets.leetcode.com/uploads/2019/09/20/1558_ex3.png
Input: n = 3, edges = [[0,1]], succProb = [0.5], start = 0, end = 2
Output: 0.00000
Explanation: There is no path between 0 and 2.
思路：
1.图形的最短路径问题要先想到 BFS，但是因为每个边的权重不同所以不能用 Queue 去
做，第一个到达终点的未必是最短路径，所以我们改用 maxHeap 来让每次都从机率最
大的路径开始计算。
2.然后额外加入一个表纪录之前算过的每个点的最大机率，只有新的机率使旧的变高才
更新。
3.如果最后到达不了 end 就返回 false。
Java Code：