1039. Minimum Score Triangulation of Polygon

最新推荐文章于 2022-10-31 11:14:40 发布

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最新推荐文章于 2022-10-31 11:14:40 发布

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分类专栏： leetcode

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博客围绕凸多边形三角剖分展开，要计算三角剖分的最小总得分，即每个三角形顶点标签乘积之和的最小值。最初从‘每个点的分配’思路着手会超时，后转换思路，考虑相邻两点即一条首尾相连的边，在求dp[s][t]时考虑该边归属来解决问题。

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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:

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

思路：一开始着手于 “每个点的分配” 上，如下：

class Solution(object):
    def minScoreTriangulation(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        memo = {}
        
        def helper(a):
            if len(a)==3: return a[0]*a[1]*a[2]
            if tuple(a) in memo: return memo[tuple(a)]
            if tuple(a[::-1]) in memo: return memo[tuple(a[::-1])]
            
            res = float('inf')
            for i in range(2,len(a)-1):
                tmp = helper(a[:i+1]) + helper(a[i:]+[a[0]])
                res = min(res, tmp)
            
            tmp = helper(a[:2]+[a[-1]]) + helper(a[1:])
            res = min(res, tmp)
            
            memo[tuple(a)] = memo[tuple(a[::-1])] = res
            return res
        
        return helper(A)

上述方法会TLE，数组转tuple耗时较多，想用(start,end)的方式表示key，但是无从下手，因为多边形是封闭的，首尾corner是连在一起的。

换一个思路是考虑相邻2个点，也就是一条边，而且这条边就是“首尾相连的边”，这样就不需要在key里面表示“首尾corner” 相连的case了，即每次在求dp[s][t]的时候就已经考虑完了“首尾相连的边”的归属，如果每次求dp[s][t]不是考虑的首尾边，就无法用一个2元组作为key

class Solution(object):
    def minScoreTriangulation(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        memo = {}
        
        def helper(a, s, t):
            if t-s+1<3: return 0
            if t-s+1==3: return a[s]*a[s+1]*a[s+2]
            if (s,t) in memo: return memo[(s,t)]
            
            res = float('inf')
            for i in range(s+1,t):
                tmp = a[s]*a[t]*a[i]
                tmp += helper(a,s,i) 
                tmp += helper(a,i,t)
                res = min(res, tmp)
            
            memo[(s,t)] = res
            return res
        
        return helper(A, 0, len(A)-1)

s=Solution()
print(s.minScoreTriangulation([1,2,3]))
print(s.minScoreTriangulation([32,43,74,29,64,98,96,19,20,33,19,7,9,98,63,6,65,31,13,97]))
print(s.minScoreTriangulation([3,7,4,5]))  
print(s.minScoreTriangulation([1,3,1,4,1,5]))