LeetCode 1039. Minimum Score Triangulation of Polygon

Posted 2020-11-14 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/minimum-score-triangulation-of-polygon/

题目：

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

题解：

Edge between A[i] and A[j] would construct only one triangle in polygon. With k between i and j, these 3 nodes construct trangle and the rest i~k, and k~j are polygons. Maintain the minimum.

Let dp[i][j] denotes the minimum score got using nodes from A[i] to A[j].

For all k bigger than i and smaller than j, maintain the mimimum score by min(dp[i][k] + dp[k][j] + A[i]*A[j]*A[k]).

Time Complexity: O(n^3). n = A.length.

Space: O(n^2).

AC Java:

 1 class Solution {
 2     public int minScoreTriangulation(int[] A) {
 3         int n = A.length;
 4         int [][] dp = new int[n][n];
 5         for(int d = 2; d<n; d++){
 6             for(int i = 0; i+d<n; i++){
 7                 int j = i+d;
 8                 dp[i][j] = Integer.MAX_VALUE;
 9                 for(int k = i+1; k<j; k++){
10                     dp[i][j] = Math.min(dp[i][j], dp[i][k]+dp[k][j]+A[i]*A[j]*A[k]);
11                 }
12             }
13         }
14         
15         return dp[0][n-1];
16     }
17 }

Another implementation.

 1 class Solution {
 2     public int minScoreTriangulation(int[] A) {
 3         int n = A.length;
 4         int [][] dp = new int[n][n];
 5         for(int j = 2; j<n; j++){
 6             for(int i = j-2; i>=0; i--){
 7                 dp[i][j] = Integer.MAX_VALUE;
 8                 for(int k = i+1; k<j; k++){
 9                     dp[i][j] = Math.min(dp[i][j], dp[i][k]+dp[k][j]+A[i]*A[j]*A[k]);
10                 }
11             }
12         }
13         
14         return dp[0][n-1];
15     }
16 }