枚举每条线段称为凸包边界的概率。注意一对点中有1个在线段左边*0.5,0个在线段左边*0,2个都在线段左边*1。最后答案要除4,因为枚举的两个点也要算概率。
#include <cstdio> #include <cstring> #include <cmath> #include <vector> #include <complex> #include <algorithm> using namespace std; typedef pair<int,int> pii; typedef long long ll; const double pi = 4 * atan(1); const double eps = 1e-10; inline int dcmp (double x) { if (fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } inline double getDistance (double x, double y) { return sqrt(x * x + y * y); } inline double torad(double deg) { return deg / 180 * pi; } struct Point { int id; double x, y, ang; Point (double x = 0, double y = 0, double ang = 0, int id = 0): x(x), y(y), ang(ang), id(id) {} void read () { scanf("%lf%lf", &x, &y); } bool operator < (const Point& u) const { return dcmp(x - u.x) < 0 || (dcmp(x-u.x)==0 && dcmp(y-u.y) < 0); } bool operator > (const Point& u) const { return u < *this; } bool operator == (const Point& u) const { return dcmp(x - u.x) == 0 && dcmp(y - u.y) == 0; } bool operator != (const Point& u) const { return !(*this == u); } bool operator <= (const Point& u) const { return *this < u || *this == u; } bool operator >= (const Point& u) const { return *this > u || *this == u; } Point operator + (const Point& u) { return Point(x + u.x, y + u.y); } Point operator - (const Point& u) { return Point(x - u.x, y - u.y); } Point operator * (const double u) { return Point(x * u, y * u); } Point operator / (const double u) { return Point(x / u, y / u); } double operator * (const Point& u) { return x*u.y - y*u.x; } }; typedef Point Vector; const int maxn = 1005; int N, V[maxn], C1, C2; Point A[maxn], B[maxn], T[maxn * 4]; double getCross (Vector a, Vector b) { return a.x * b.y - a.y * b.x; } double getArea (Point a, Point b, Point c) { return getCross(b - a, c - a) / 2; } inline int cmp(const Point& a, const Point& b) { return a.ang < b.ang; } void add (int x) { V[x]++; if (V[x] == 1) C1++; else C1--, C2++; } void del (int x) { V[x]--; if (V[x] == 0) C1--; else C2--, C1++; } double solve(Point* P) { double ans = 0; for (int i = 0; i < N; i++) { int sz = 0; for (int j = 0; j < N; j++) if (i != j) { T[sz++] = Point(A[j].x, A[j].y, atan2(A[j].y-P[i].y, A[j].x-P[i].x), j); T[sz++] = Point(B[j].x, B[j].y, atan2(B[j].y-P[i].y, B[j].x-P[i].x), j); } for (int j = 0; j < sz; j++) { T[sz + j] = T[j]; T[sz + j].ang += 2 * pi; } sort(T, T + 2 * sz, cmp); int r = 0; C1 = C2 = 0; memset(V, 0, sizeof(V)); add(T[r].id); for (int j = 0; j < sz; j++) { del(T[j].id); while (T[r+1].ang - T[j].ang < pi) add(T[++r].id); //while (C1 + 2 * C2 != r - j); int c = C1 - V[T[j].id]; if (c + C2 != N - 2) continue; ans += getArea(Point(0, 0), P[i], T[j]) * pow(0.5, c); } } return ans; } int main () { while (scanf("%d", &N) == 1) { for (int i = 0; i < N; i++) A[i].read(), B[i].read(); double ans = solve(A) + solve(B); printf("%lf\n", ans / 4); } return 0; }
版权声明:本文为博主原创文章,未经博主允许不得转载。
时间: 2024-10-29 19:09:54