### PROBLEM LINK:

**Author:** Praveen Dhinwa

**Tester:** Utkarsh Lath

**Editorialist:** Balajiganapathi Senthilnathan

**Russian Translator:** Sergey Kulik

**Mandarian Translator:** Gedi Zheng

### DIFFICULTY:

Hard

### PREREQUISITES:

Geometry

### PROBLEM:

Given n points each with a positive weight, find the minimum weighted convex polygon with k vertices chosen among the n points. Do this for all k from 3 to n. The weight of a polygon is the sum of the weights of the points enclosed by it (including the boundaries).

### SHORT EXPLANATION

Suppose we fix a point p_1 as the lower right corner of the polygon we are about to construct. We will sort all the valid points (i.e. those that are above p_1) in counter clock wise direction with respect to p_1. Now let us define dp(p_3, p_2, k) as the minimum weighted polygon which has p_1 as the lowest point, p_3 and p_2 are the last two added points and we have to add k more vertices. Now, for all vertices in counter clock wise order, we try adding each point i which is to the left of the line connecting p2 to p3 and recursively solve the dp(i, p3, k + 1) and add we also add the weights of the point inside the triangle p2 - p3 - i. We take minimum over all such i and that will be the answer for dp(p_3, p_2, k).

Now we try fixing each of the point as the lower most one (i.e. p_1) and do the above procedure. For each k, we take minimum over all such fixed points. We do this for all k (3 \leq k \leq n) to get the final answer$.

### EXPLANATION:

Let us fix a point p_1 as the bottom most point of the polygon (i.e. the one with the least y coordinate). Since this is the bottom most point, we can ignore all the points which have y values less than p_1. Now we sort the remaining points in counter clockwise direction. We want to get the minimum weighted convex polygon for each k having p_1 as the bottom most point.

The key observation is that suppose we have a minimum weighted convex polygon with k vertices, then if we remove any vertex, we will be left with a minimum weighted convex polygon of k - 1 vertices having p_1 as the bottom most point. So, to get the answer for k, we try adding each valid point as the next vertex and solve the problem for k - 1 vertices. Then take the minimum over all such valid points to get the minimum value for k vertices.

In other words, we are building the polygon triangle by triangle. What other information are needed for implementing this?

- We need to know the last two added points added to the polygon. This is to ensure that the next point we add does make a convex polygon (i.e it has to be to the left of the line made by the last two points).
- We need to know the weight of all possible triangles. We can pre-calculate this in O(n^4) time and O(n^3) space. Let us define totalWeightsInside[i][j][k] as the weight of the points inside the triangle formed by points i, j and k. We can calculate this by looping through all points and adding its weight to totalWeightsInside[i][j][k] if that point lies inside the triangle formed by i, j and k.

Let the function dp(p_3, p_2, k) return the minimum weighted convex polygon with p_1 as the bottom most vertex, p_3 and p_2 as the last two vertex and k vertices remaining to be added. Now we loop through all valid points i which are ahead of p_3 in counter clockwise order and see if can add i to the polygon. We can add i only if it is to the left of the line formed by p_2 - p_3. Otherwise the angle made by p_2 - p_3 - i will be greater than 180 degrees and hence it won’t be a convex polygon. Now to get the weight, we recursively solve for dp(i, p3, k - 1) and add the weights of the points inside the triangle formed by p_2 - p_3 - i to the weight. We try this for all i and the minimum among them would be the answer. Finally note that for a fixed p_1, the state may repeat. So, we memoize the values to avoid recalculation.

For getting the final answer, we loop through all points and fix them as p_1. Then we store all the points which are not below p_1 and sort them in counter clockwise order. Now we have to select the next 2 vertices of the polygon. So we loop through all pairs of valid points and keep them as the next 2 vertices. Finally we need to solve for all k, so we will loop for k from 3 to n and call dp(p3, p2, k - 3) to get the answer for this case. We keep updating the minimum weighted polygon for k vertices using this value.

Please consult author’s solution for the sample implementation of this idea.

#### Time Complexity:

\mathcal{O} (n^5)

- \mathcal{O}(n) for fixing vertex p_1
- dp(p_3, p_2, k) has total number of states equal to \mathcal{O}(n^3) and total number of transitions equal to \mathcal{O}(n). Hence, time taken will be \mathcal{O}(n^4).
- Finally, the total time will be product of both, i.e. \mathcal{O}(n^5)