[Swift]LeetCode1039. 多边形三角剖分的最低得分 | Minimum Score Triangulation of Polygon

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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:

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

---

给定 N，想象一个凸 N 边多边形，其顶点按顺时针顺序依次标记为 A[0], A[i], ..., A[N-1]。

假设您将多边形剖分为 N-2 个三角形。对于每一个三角形，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全部 N-2 个三角形的值之和。

返回多边形进行三角剖分后能够获得的最低分。

示例 1：

输入：[1,2,3]
输出：6
解释：多边形已经三角化，惟一三角形的分数为 6。

示例 2：

输入：[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。

示例 3：

输入：[1,3,1,4,1,5]
输出：13
解释：最低分数三角剖分的得分状况为 1*1*3 + 1*1*4 + 1*1*5 + 1*1*1 = 13。

提示：

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

---

20ms

 1 class Solution {
 2     func minScoreTriangulation(_ A: [Int]) -> Int {
 3         var dp = Array(repeating: Array(repeating: 0, count: A.count), count: A.count + 1)
 4         for size in 3...A.count {
 5             for i in 0...A.count - size {
 6                 if size == 3 {
 7                     dp[size][i] = A[i] * A[i + 1] * A[i + 2]
 8                 } else {
 9                     var result = Int.max
10                     
11                     for other in (i + 1)..<(i + size - 1) {
12                         var this = A[i] * A[i + size - 1] * A[other]
13                         
14                         if other > i + 1 {
15                             this += dp[other - i + 1][i]
16                         }
17                         
18                         if other < i + size - 2 {
19                             this += dp[i + size - other][other]
20                         }
21                         
22                         result = min(result, this)
23                     }
24                     
25                     dp[size][i] = result
26                 }
27             }
28         }
29         
30         return dp[A.count][0]
31     }
32 }

---

Runtime: 36 ms

Memory Usage: 18.8 MB

 1 class Solution {
 2     func minScoreTriangulation(_ A: [Int]) -> Int {
 3         var n:Int = A.count
 4         var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n),count:n)
 5         for len in 2..<n
 6         {
 7             for i in 0..<(n - len)
 8             {
 9                 var j:Int = i + len
10                 dp[i][j] = Int.max
11                 for k in (i + 1)..<j
12                 {
13                     dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + A[i]*A[j]*A[k])
14                 }
15             }
16         }
17         return dp[0][n-1]        
18     }
19 }

---

44ms

 

 1 class Solution {
 2     func minScoreTriangulation(_ A: [Int]) -> Int {
 3         var memo = [[Int]](repeating: [Int](repeating: 0, count: A.count), count:A.count)
 4         return dfsHelper(A, 0, A.count - 1, &memo)
 5     }
 6     
 7     fileprivate func dfsHelper(_ A:[Int], _ i:Int, _ j: Int, _ memo: inout [[Int]]) -> Int {
 8         
 9         if j < i + 2 {
10             return 0 
11         }
12         if memo[i][j] != 0 {
13             return memo[i][j]
14         }
15         var result = Int.max
16         for k in stride(from: i+1, to: j, by: 1) {
17             result = min(result, dfsHelper(A, i, k, &memo) + dfsHelper(A, k, j, &memo) + A[i] * A[k] * A[j])
18         }
19         memo[i][j] = result
20         return memo[i][j]
21     }
22 }