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+// 1039. Minimum Score Triangulation of Polygon
+#include <limits.h>
+#include <math.h>
+#include <stdio.h>
+#include <stdlib.h>
+
+/*
+ * given a convex n-sided polygon where each vertex has an integer value. you
+ * are given an integer array 'values' where 'values[i]' is the value of the
+ * i'th vertex (ie. clockwise order). you will triangulate the polygon into 'n -
+ * 2' triangles. for each triangle, the value of that triangle is the product of
+ * the values of its 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
+ */
+
+int minScoreTriangulation(int *values, int values_size) {
+ int n = values_size, **dp = malloc(n * sizeof(int *));
+ for (int i = 0; i < n; i++) {
+ dp[i] = malloc(n * sizeof(int));
+ dp[i][i] = 0;
+ if (i + 1 < n)
+ dp[i][i + 1] = 0;
+ for (int j = i + 2; j < n; j++)
+ dp[i][j] = INT_MAX / 3;
+ }
+ for (int len = 3; len <= n; len++) {
+ for (int i = 0; i < n; i++) {
+ int j = i + len - 1;
+ if (j >= n)
+ break;
+ for (int k = i + 1; k < j; k++)
+ dp[i][j] = fmin(dp[i][j], dp[i][k] + values[i] * values[k] * values[j] +
+ dp[k][j]);
+ }
+ }
+ int ans = dp[0][n - 1];
+ for (int i = 0; i < n; i++)
+ free(dp[i]);
+ free(dp);
+ return ans;
+}
+
+int main() {
+ int v1[] = {1, 2, 3}, v2[] = {3, 7, 4, 5}, v3[] = {1, 3, 1, 4, 1, 5};
+ printf("%d\n", minScoreTriangulation(v1, 3)); // expect: 6
+ printf("%d\n", minScoreTriangulation(v2, 4)); // expect: 144
+ printf("%d\n", minScoreTriangulation(v3, 6)); // expect: 13
+}