多边形三角剖分的最小值 python

v[i][j]
	从顶点i到j组成的凸多边形
	边界
		v[i][i] = 0;
		v[i][i + 1] = 0
		v[i][i + 2] = A[i] * A[i + 1] * A[i + 2]

当多边形是三条边以上时(i + 2 < j)
	v[i, j]可以通过一个三角形(i, j, k)划分为两部分
		凸多边形v[i, k]
		凸多边形v[k, j]
i.e:
	v[i, j] = MIN(v[i][k] + v[k][j] + A[i] * A[k] * A[j]) (i < k < j)



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from typing import List
from functools import lru_cache


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        @lru_cache(None)
        def v(i, j):
        	# recursive end
            if i + 1 == j:
                return 0
                
            res = float('inf')
            # search k that results in minum value
            for k in range(i + 1, j):
                res = min(res, v(i, k) + v(k, j) + A[i] * A[j] * A[k])
            return res

        return v(0, len(A) - 1)


if __name__ == "__main__":
    a = Solution()
    print(a.minScoreTriangulation([35,73,90,27,71,80,21,33,33,13,48,12,68,70,80,36,66,3,70,58]))


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