[Leetcode 1039] Minimum Score Triangulation of Polygon

原题说明

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: 375 + 457 = 245, or 345 + 347 = 144.  The minimum score is 144.

Example 3:

Input: [1,3,1,4,1,5]
Output: 13
Explanation: The minimum score triangulation has score 113 + 114 + 115 + 111 = 13.

Note:

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

解题思路

动态规划，递归可以使逻辑简单（本质还是动态规划）

将多边形起始位置设为start，end, 用一个数组dp来记录任意起始位置的score
为了计算dp[start][end], 我们用一个index k在start到end之间遍历
dp[start][end] = min(dp[start][k] + dp[k][end] + A[start] * A[k] * A[end])
结果为dp[0][n - 1]
注意：
相邻的dp[i][i + 1] = 0, 因为两条边无法组成三角形
如果用传统动归的方法，必须i的位置从j开始往前推，不能i和j都从小往大推，不然dp[k][j]是未知的。

示例代码 (cpp)

递归（实际上还是动态规划，因为有memorization）

class Solution {
    int getScore(const vector<int>& A, vector<vector<int>>& dp, int start, int end) {
        if (start >= end - 1) {
            return 0;
        }
        if (dp[start][end] > 0) {
            return dp[start][end];
        }
        int score = INT_MAX;
        for (int k = start + 1; k <= end - 1; ++k) {
            score = min(score, getScore(A, dp, start, k) + getScore(A, dp, k, end) + A[start] * A[k] * A[end]);
        }
        dp[start][end] = score;
        return score;
    }
public:
    // 递归（实际上还是动态规划，因为有memorization）
    int minScoreTriangulation(vector<int>& A) {
        int n = A.size();
        vector<vector<int>> dp(n, vector<int>(n, 0));
        return getScore(A, dp, 0, n - 1);
    }
};

动态规划

class Solution {
public:
    int minScoreTriangulation(vector<int>& A) {
        int n = A.size();
        vector<vector<int>> dp(n, vector<int>(n, 0));
        for (int j = 2; j < n; ++j) {
            for (int i = j - 2; i >= 0; --i) {
                dp[i][j] = INT_MAX;
                for (int k = i + 1; k < j; ++k) {
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + A[i] * A[j] * A[k]);
                }
            }
        }
        return dp[0][n - 1];
    } 
};

示例代码 (java)

class Solution {
    public int minScoreTriangulation(int[] A) {
        int n = A.length;
        int[][] dp = new int[n][n];
        for (int j = 2; j < n; ++j) {
            for (int i = j - 2; i >= 0; --i) {
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i + 1; k < j; ++k)
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k][j] + A[i] * A[j] * A[k]);
            }
        }
        return dp[0][n - 1];
    }
}

示例代码 (python)

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        n = len(A)
        dp = [[0] * n for i in range(n)]
        for d in range(2, n):
            for i in range(n - d):
                j = i + d
                dp[i][j] = min(dp[i][k] + dp[k][j] + A[i] * A[j] * A[k] for k in range(i + 1, j))
        return dp[0][n - 1]

复杂度分析

时间复杂度:O(N^3), 用传统动归可以很容易看出，用递归的方法可以想到任意dp[i][j]都遍历一次O(N^2)，而每一次遍历中
都有一个 for (int k = start + 1; k <= end - 1; ++k)用到O(N)
空间复杂度:O(N^2), 递归会使用额外的O(N)stack

归纳总结

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