5047. Minimum Score Triangulation of Polygon-优快云博客

本文链接：https://blog.youkuaiyun.com/EasonDongH/article/details/89842496

博客围绕凸N边形三角剖分展开，将其划分为N - 2个三角形，每个三角形的值为顶点标签之积，三角剖分总分是所有三角形值之和。通过多个示例给出不同顶点标签数组下的输入输出，要求计算最小总分，还给出了数组长度和元素值范围。

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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: 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

class Solution {
public:
    static const int MAX = 55;
    static const int INF = 1e9 + 5;

    int n;
    vector<int> A;
    int dp[MAX][MAX];

    int prev(int x) {
        return (x + n - 1) % n;
    }

    int next(int x) {
        return (x + 1) % n;
    }

    int solve(int start, int end) {
        if (start == end || next(start) == end || prev(start) == end)
            return 0;

        int &answer = dp[start][end];

        if (answer >= 0)
            return answer;

        answer = INF;

        for (int i = next(start); i != end; i = next(i))
            answer = min(answer, A[start] * A[i] * A[end] + solve(start, i) + solve(i, end));

        return answer;
    }

    int minScoreTriangulation(vector<int>& _A) {
        A = _A;
        n = A.size();
        memset(dp, -1, sizeof(dp));
        int best = INF;

        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                for (int k = j + 1; k < n; k++)
                    best = min(best, A[i] * A[j] * A[k] + solve(i, j) + solve(j, k) + solve(k, i));

        return best;
    }
};