1039.minimum-score-triangulation-of-polygon.cpp

/*
 * @lc app=leetcode id=1039 lang=cpp
 *
 * [1039] Minimum Score Triangulation of Polygon
 *
 * https://leetcode.com/problems/minimum-score-triangulation-of-polygon/description/
 *
 * algorithms
 * Medium (42.29%)
 * Total Accepted:    3.6K
 * Total Submissions: 8.6K
 * Testcase Example:  '[1,2,3]'
 *
 * 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
 * 
 * 
 * 
 * 
 */
class Solution {
public:
    int n;
    int dp[50][50];
    int callme(vector<int>& A, int i, int j){
        if(j+1 == i)
            return 0;
        if(j+2 == i){
            return A[i]*A[j]*A[j+1];
        }
        // if(mm.find(i)!=mm.end() && mm[i].find(j)!=mm[i].end())
        //     return mm[i][j];
        if(dp[i][j]!=-1)
            return dp[i][j];
        int res = INT_MAX;
        for(int k = j+1; k!=i; k++){
            int tmp = callme(A, k, j) + A[i]*A[k]*A[j] + callme(A, i, k);
            res = min(tmp, res);
        }
        dp[i][j] = res;
        return res;
    }
    int minScoreTriangulation(vector<int>& A) {
        n = A.size();
        // mm.clear();
        for(int i=0; i<50; i++)
            for(int j=0; j<50; j++)
                dp[i][j] = -1;
        return callme(A, n-1, 0);

    }
};