LA 4064 (计数 极角排序) Magnetic Train Tracks

这个题和UVa11529很相似。

枚举一个中心点，然后按极角排序，统计以这个点为钝角的三角形的个数，然后用C(n, 3)减去就是答案。

另外遇到直角三角形的情况很是蛋疼，可以用一个eps，不嫌麻烦的话就用整数的向量做点积。

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long LL;
const int maxn = 1200 + 10;
const double PI = acos(-1.0);
const double eps = 1e-9;

struct Point
{
    double x, y;
    Point() {}
    Point(double x, double y):x(x), y(y) {}
}p[maxn], p2[maxn];

Point operator - (const Point& A, const Point& B)
{
    return Point(A.x-B.x, A.y-B.y);
}

double ang[maxn * 2];

LL inline C3(int n) { return (LL)n * (n-1) / 2 * (n-2) / 3; }

int main()
{
    //freopen("in.txt", "r", stdin);
    int n, kase = 0;

    while(scanf("%d", &n) == 1 && n)
    {
        for(int i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);

        LL cnt = 0;
        for(int i = 0; i < n; i++)
        {
            int k = 0;
            for(int j = 0; j < n; j++) if(j != i)
            {
                p2[k] = p[j] - p[i];
                ang[k] = atan2(p2[k].y, p2[k].x);
                ang[k + n - 1] = ang[k] + PI * 2.0;
                k++;
            }
            k = 2*n-2;
            sort(ang, ang + k);
            int L, R1 = 0, R2 = 0;
            for(L = 0; L < n-1; L++)
            {
                double b1 = ang[L] + PI / 2;
                double b2 = ang[L] + PI;
                while(ang[R1] <= b1 - eps) R1++;
                while(ang[R2] < b2) R2++;
                cnt += R2 - R1;
            }
        }
        LL ans = C3(n) - cnt;
        printf("Scenario %d:
There are %lld sites for making valid tracks
", ++kase, ans);
    }

    return 0;
}