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"translatedContent": "<p>给你一个由 <code>n</code> 个节点(下标从 0 开始)组成的无向加权图,该图由一个描述边的列表组成,其中 <code>edges[i] = [a, b]</code> 表示连接节点 a 和 b 的一条无向边,且该边遍历成功的概率为 <code>succProb[i]</code> 。</p>\n\n<p>指定两个节点分别作为起点 <code>start</code> 和终点 <code>end</code> ,请你找出从起点到终点成功概率最大的路径,并返回其成功概率。</p>\n\n<p>如果不存在从 <code>start</code> 到 <code>end</code> 的路径,请 <strong>返回 0</strong> 。只要答案与标准答案的误差不超过 <strong>1e-5 </strong>,就会被视作正确答案。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/07/12/1558_ex1.png\" style=\"height: 186px; width: 187px;\"></strong></p>\n\n<pre><strong>输入:</strong>n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start = 0, end = 2\n<strong>输出:</strong>0.25000\n<strong>解释:</strong>从起点到终点有两条路径,其中一条的成功概率为 0.2 ,而另一条为 0.5 * 0.5 = 0.25\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/07/12/1558_ex2.png\" style=\"height: 186px; width: 189px;\"></strong></p>\n\n<pre><strong>输入:</strong>n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.3], start = 0, end = 2\n<strong>输出:</strong>0.30000\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/07/12/1558_ex3.png\" style=\"height: 191px; width: 215px;\"></strong></p>\n\n<pre><strong>输入:</strong>n = 3, edges = [[0,1]], succProb = [0.5], start = 0, end = 2\n<strong>输出:</strong>0.00000\n<strong>解释:</strong>节点 0 和 节点 2 之间不存在路径\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 10^4</code></li>\n\t<li><code>0 &lt;= start, end &lt; n</code></li>\n\t<li><code>start != end</code></li>\n\t<li><code>0 &lt;= a, b &lt; n</code></li>\n\t<li><code>a != b</code></li>\n\t<li><code>0 &lt;= succProb.length == edges.length &lt;= 2*10^4</code></li>\n\t<li><code>0 &lt;= succProb[i] &lt;= 1</code></li>\n\t<li>每两个节点之间最多有一条边</li>\n</ul>\n",
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"hints": [
"Multiplying probabilities will result in precision errors.",
"Take log probabilities to sum up numbers instead of multiplying them.",