<p>You are given an array <code>trees</code> where <code>trees[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents the location of a tree in the garden.</p>

<p>Fence the entire garden using the minimum length of rope, as it is expensive. The garden is well-fenced only if <strong>all the trees are enclosed</strong>.</p>

<p>Return <em>the coordinates of trees that are exactly located on the fence perimeter</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/erect2-plane.jpg" style="width: 400px; height: 393px;" />
<pre>
<strong>Input:</strong> trees = [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]]
<strong>Output:</strong> [[1,1],[2,0],[4,2],[3,3],[2,4]]
<strong>Explanation:</strong> All the trees will be on the perimeter of the fence except the tree at [2, 2], which will be inside the fence.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/erect1-plane.jpg" style="width: 400px; height: 393px;" />
<pre>
<strong>Input:</strong> trees = [[1,2],[2,2],[4,2]]
<strong>Output:</strong> [[4,2],[2,2],[1,2]]
<strong>Explanation:</strong> The fence forms a line that passes through all the trees.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= trees.length &lt;= 3000</code></li>
	<li><code>trees[i].length == 2</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 100</code></li>
	<li>All the given positions are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    static bool mycmp(const vector<int> &a, const vector<int> &b) {
        return a[0] < b[0];
    }
    
    int orientation(vector<int> &p, vector<int> &q, vector<int> &r ){
        int value = (q[0]-p[0])*(r[1]-p[1]) - (r[0]-p[0])*(q[1]-p[1]);
        if(value == 0) return 0;
        return (value>0)?1:-1;
    }
    
    double distPointLine(const vector<int> &A, const vector<int> &B, const vector<int> &C) {
        // Distance from point C to line AB
        int a = B[1] - A[1], b = B[0] - A[0];
        return abs((a * C[0] - b * C[1] + B[0] * A[1] - B[1] * A[0]) / sqrt(a * a + b * b));
    }
    
    void FindHull(vector<vector<int>> &points, vector<int> &A, vector<int> &B) {
        if (points.empty()) return;
        
        int idx = 0;
        double dist = distPointLine(A, B, points[0]);
        for (int i = 1; i < points.size(); i++) {
            if (distPointLine(A, B, points[i]) > dist) {
                dist = distPointLine(A, B, points[i]);
                idx = i;
            }
        }
        ret.push_back(points[idx]);
        
        vector<vector<int>> R, T;
        for (int i = 0; i < points.size(); i++) {
            if (i != idx) {
                int tmp = orientation(A, points[idx], points[i]);
                if (tmp >= 0) R.push_back(points[i]);
                else {
                    tmp = orientation(points[idx], B, points[i]);
                    if (tmp >= 0) T.push_back(points[i]);
                }
            }
        }
        FindHull(R, A, points[idx]);
        FindHull(T, points[idx], B);
    }
    
    vector<vector<int>> ret;
    
    vector<vector<int>> outerTrees(vector<vector<int>>& points) {
        if (points.size() <= 1) return points;
        
        sort(points.begin(), points.end(), mycmp);
        ret.push_back(points[0]);
        ret.push_back(points.back());
        
        vector<vector<int>> Left, Right, Online;
        for (int i = 1; i < points.size() - 1; i++) {
            int tmp = orientation(points[0], points.back(), points[i]);
            if (tmp < 0){
                Right.push_back(points[i]);
            }
            else if (tmp > 0){
                Left.push_back(points[i]);
            }
            else{
                Online.push_back(points[i]);
            }
        }
        
        if (Left.empty() || Right.empty()) {
            for (int i = 0; i < Online.size(); i++){
                ret.push_back(Online[i]);
            }
        }

        FindHull(Left, points[0], points.back());
        FindHull(Right, points.back(), points[0]);
        
        return ret;
    }
};

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