https://leetcode.com/problems/path-with-maximum-probability/description/
1514. Path with Maximum Probability
給你一個大小為 n 的無向圖資訊，陣列 edges[] 表示邊關係，succProb[] 表示邊的機
率權重，路徑經過時需要乘以該權重，求出從 start 走到 end 的路徑，他的最終權重必
需是最高，走不到則返回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.圖形的最短路徑問題要先想到 BFS，但是因為每個邊的權重不同所以不能用 Queue 去
做，第一個到達終點的未必是最短路徑，所以我們改用 maxHeap 來讓每次都從機率最
大的路徑開始計算。
2.然後額外加入一個表紀錄之前算過的每個點的最大機率，只有新的機率使舊的變高才
更新。
3.如果最後到達不了 end 就返回 false。
Java Code：