How Many Triangles

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5784

Description

Alice has n points in two-dimensional plane. She wants to know how many different acute triangles they can form. Two triangles are considered different if they differ in at least one point.

Input

The input contains multiple test cases.
For each test case, begin with an integer n,
next n lines each contains two integers xi and yi.
3≤n≤2000
0≤xi,yi≤1e9
Any two points will not coincide.

Output

For each test case output a line contains an integer.

Sample Input

3
1 1
2 2
2 3
3
1 1
2 3
3 2
4
1 1
3 1
4 1
2 3

Sample Output

0
1
2

Hint

题意

平面给你2000个不重合的点，问你有多少个锐角三角形

题解：

数一数锐角的数量A和直角+钝角的数量B，那么答案就是(A-2B)/3。 暴力算的话是\(O(n^3)\)的。使用极角排序+two pointers就可以做到\(O(n^2log\ n)\)
这边钝角指代范围在90度到180度之间的角(不包括90和180)。

我们枚举一个点，算出所有向量，然后枚举一个向量，towpointer很容易算出直角那条线，和钝角那条线，然后就可以统计个数了。

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2005;
const int Q = 1e9 + 7;

struct Point {
    int x , y;
    Point (int _x = 0 , int _y = 0) {
        x = _x , y = _y;
    }
    Point operator - (const Point &R) const {
        return Point(x - R.x , y - R.y);
    }
    LL operator ^ (const Point &R) const {
        return (LL)x * R.y - (LL)y * R.x;
    }
    LL operator % (const Point &R) const {
        return (LL)x * R.x + (LL)y * R.y;
    }
    bool sign() const {
        return y > 0 || (y == 0 && x > 0);
    }
    bool operator < (const Point &R) const {
        if (sign() != R.sign()) {
            return sign() > R.sign();
        }
        return (*this ^ R) > 0;
    }
};
int n;
Point P[N];

void work() {
    for (int i = 0 ; i < n ; ++ i) {
        scanf("%d%d" , &P[i].x , &P[i].y);
    }
    LL res = 0;
    for (int i = 0 ; i < n ; ++ i) {
        vector<Point> V;
        for (int j = 0 ; j < n ; ++ j) {
            if (P[j].x != P[i].x || P[j].y != P[i].y)
                V.push_back(P[j] - P[i]);
        }

        sort(V.begin() , V.end());
        int m = V.size();
        int e = 0 , p = 0 , w = 0;
        for (int j = 0 ; j < m ; ++ j) {
            while (e < m && (V[j] ^ V[(j + e) % m]) == 0) {
                ++ e;
            }
            p = max(e , p);
            while (p < m && ((V[j] ^ V[(j + p) % m]) > 0 && (V[j] % V[(j + p) % m]) > 0)) {
                ++ p;
            }
            w = max(w , p);
            while (w < m && ((V[j] ^ V[(j + w) % m]) > 0 && (V[j] % V[(j + w) % m]) <= 0)) {

                ++ w;
            }
            res += p - e;
            res -= 2 * (w - p);
            e = max(1 , e - 1);
            p = max(1 , p - 1);
            w = max(1 , w - 1);
        }
    }
    cout << res / 3 << endl;
}

int main() {
    while (~scanf("%d" , &n)) {
        work();
    }
    return 0;
}