minimum-score-triangulation-of-polygon

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Algorithms

You have a convex <code>n</code>-sided polygon where each vertex has an integer value. You are given an integer array <code>values</code> where <code>values[i]</code> is the value of the <code>ith</code> vertex in clockwise order.

Polygon triangulation is a process where you divide a polygon into a set of triangles and the vertices of each triangle must also be vertices of the original polygon. Note that no other shapes other than triangles are allowed in the division. This process will result in <code>n - 2</code> triangles.

You will triangulate the polygon. For each triangle, the weight of that triangle is the product of the values at its vertices. The total score of the triangulation is the sum of these weights over all <code>n - 2</code> triangles.

Return the minimum possible score that you can achieve with some triangulation of the polygon.


&nbsp;
Example 1:

<img alt="" src="http://127.0.0.1:49174/shape1.jpg" />

<div class="example-block">
Input: values = [1,2,3]

Output: 6

Explanation: The polygon is already triangulated, and the score of the only triangle is 6.
</div>

Example 2:

<img alt="" src="http://127.0.0.1:49174/shape2.jpg" />

<div class="example-block">
Input: values = [3,7,4,5]

Output: 144

Explanation: There are two triangulations, with possible scores: 3*7*5 + 4*5*7 = 245, or 3*4*5 + 3*4*7 = 144. 
The minimum score is 144.
</div>

Example 3:

<img alt="" src="http://127.0.0.1:49174/shape3.jpg" />

<div class="example-block">
Input: values = [1,3,1,4,1,5]

Output: 13

Explanation: The minimum score triangulation is 1*1*3 + 1*1*4 + 1*1*5 + 1*1*1 = 13.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>n == values.length</code></li>
	<li><code>3 &lt;= n &lt;= 50</code></li>
	<li><code>1 &lt;= values[i] &lt;= 100</code></li>
</ul>

minScoreTriangulation

Minimum Score Triangulation of Polygon

Array

Dynamic Programming

你有一个凸的<meta charset="UTF-8" />&nbsp;<code>n</code>&nbsp;边形，其每个顶点都有一个整数值。给定一个整数数组<meta charset="UTF-8" />&nbsp;<code>values</code>&nbsp;，其中<meta charset="UTF-8" />&nbsp;<code>values[i]</code>&nbsp;是第 <code>i</code> 个顶点的值（即 顺时针顺序 ）。

假设将多边形 剖分&nbsp;为 <code>n - 2</code>&nbsp;个三角形。对于每个三角形，该三角形的值是顶点标记的乘积，三角剖分的分数是进行三角剖分后所有 <code>n - 2</code>&nbsp;个三角形的值之和。

返回 多边形进行三角剖分后可以得到的最低分 。 
&nbsp;

<ol>
</ol>

示例 1：

<img alt="" src="https://assets.leetcode.com/uploads/2021/02/25/shape1.jpg" />

<pre>
输入：values = [1,2,3]
输出：6
解释：多边形已经三角化，唯一三角形的分数为 6。
</pre>

示例 2：

<img alt="" src="https://assets.leetcode.com/uploads/2021/02/25/shape2.jpg" style="height: 163px; width: 446px;" />

<pre>
输入：values = [3,7,4,5]
输出：144
解释：有两种三角剖分，可能得分分别为：3*7*5 + 4*5*7 = 245，或 3*4*5 + 3*4*7 = 144。最低分数为 144。
</pre>

示例 3：

<img alt="" src="https://assets.leetcode.com/uploads/2021/02/25/shape3.jpg" />

<pre>
输入：values = [1,3,1,4,1,5]
输出：13
解释：最低分数三角剖分的得分情况为 1*1*3 + 1*1*4 + 1*1*5 + 1*1*1 = 13。
</pre>

&nbsp;

提示：

<ul>
	<li><code>n == values.length</code></li>
	<li><code>3 &lt;= n &lt;= 50</code></li>
	<li><code>1 &lt;= values[i] &lt;= 100</code></li>
</ul>