leetcode1039. Minimum Score Triangulation of Polygon

Posted 2021-12-13 seyjs

题目如下：

Given N, consider a convex N-sided polygon with vertices labelled A[0], A[i], ..., A[N-1] in clockwise order.

Suppose you triangulate the polygon into N-2 triangles.  For each triangle, the value of that triangle is the product of the labels of the vertices, and the total score of the triangulation is the sum of these values over all N-2 triangles in the triangulation.

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

 

Example 1:

Input: [1,2,3]
Output: 6
Explanation: The polygon is already triangulated, and the score of the only triangle is 6.

Example 2:

Input: [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.

Example 3:

Input: [1,3,1,4,1,5]
Output: 13
Explanation: The minimum score triangulation has score 1*1*3 + 1*1*4 + 1*1*5 + 1*1*1 = 13.

 

Note:

1. 3 <= A.length <= 50
2. 1 <= A[i] <= 100

解题思路：这里推荐一本书《趣学算法》，里面有几个专题，讲解也非常有意思。本题对应书中的4.7章：最优三角剖分，解答如下图。

代码如下：

class Solution(object):
    def minScoreTriangulation(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        dp = []
        for i in A:
            dp.append([0] * len(A))
        # dp[i][j] = dp[i][k] + dp[k+1][j] + A[i]+A[j]+A[k]
        for i in range(len(A)-3,-1,-1):
            for j in range(i+2,len(A)):
                for k in range(i+1,j):
                    if dp[i][j] == 0 or dp[i][j] > dp[i][k] + dp[k][j] + A[i]*A[j]*A[k]:
                        dp[i][j] = dp[i][k] + dp[k][j] + A[i]*A[j]*A[k]
        #print dp
        return dp[0][-1]