UVA 11186 - Circum Triangle(计算几何+容斥)

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这题用n^2的算法能过，先任意枚举两点，和圆心组成的三角形求面积，这个面积可能会被加(n - 2)次，但是要注意，如果有3点是在同一侧，那么要减去，于是在枚举一遍，每次枚举一个点，然后枚举和这个点度数相差180以内的点，求面积，这个面积要减去2 * （j - i + 1)次

代码：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int N = 505;const double PI = acos(-1.0);int n;double r, d[N * 2];struct Point {    double x, y;    Point() {}    Point(double x, double y) {        this->x = x;        this->y = y;    }};typedef Point Vector;Vector operator - (Vector A, Vector B) {    return Vector(A.x - B.x, A.y - B.y);}inline double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积inline double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积inline double Are(Point A, Point B, Point C) {return fabs(Area2(A, B, C)) / 2;}inline Point getP(double d) {    double rad = d / 180 * PI;    return Point(cos(rad) * r, sin(rad) * r);}int main() {    while (~scanf("%d%lf", &n, &r) && n || r) {        for (int i = 0; i < n; i++) scanf("%lf", &d[i]);        sort(d, d + n);        double ans = 0;        for (int i = 0; i < n; i++)            for (int j = i + 1; j < n; j++)                ans += Are(Point(0, 0), getP(d[i]), getP(d[j])) * (n - 2);        for (int i = 0; i < n; i++)            d[i + n] = d[i] + 360;        for (int i = 0; i < n; i++) {            for (int j = i + 2; d[j] - d[i] < 180; j++) {                ans -= 2 * Are(Point(0, 0), getP(d[i]), getP(d[j % n])) * (j - i - 1);            }        }        printf("%.0f\n", ans);    }    return 0;}

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