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Union resulting in two holes with shared node yields invalid result
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AtCoder 3868 Holes（Graham扫描法）
20180325 22:48:46B  HolesTime limit : 2sec / Memory limit : 256MBScore : 600 pointsProblem StatementThere are N holes in a twodimensional plane. The coordinates of the ith hole are (xi,yi).Let R=10101010....B  Holes
Time limit : 2sec / Memory limit : 256MB
Score : 600 points
Problem Statement
There are N holes in a twodimensional plane. The coordinates of the ith hole are (xi,yi).
Let R=10101010. Ringo performs the following operation:
 Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.
For every i (1≤i≤N), find the probability that Snuke falls into the ith hole.
Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:
 Pick two real numbers x and y independently according to uniform distribution on [−R,R].
 If x2+y2≤R2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.
Constraints
 2≤N≤100
 xi,yi≤106(1≤i≤N)
 All given points are pairwise distinct.
 All input values are integers.
Input
Input is given from Standard Input in the following format:
N x1 y1 : xN yN
Output
Print N real numbers. The ith real number must represent the probability that Snuke falls into the ith hole.
The output will be judged correct when, for all output values, the absolute or relative error is at most 10−5.
Sample Input 1
Copy2 0 0 1 1
Sample Output 1
Copy0.5 0.5
If Ringo put Snuke in the region x+y≤1, Snuke will fall into the first hole. The probability of this happening is very close to 0.5. Otherwise, Snuke will fall into the second hole, the probability of which happening is also very close to 0.5.
Sample Input 2
Copy5 0 0 2 8 4 5 2 6 3 10
Sample Output 2
Copy0.43160120892732328768 0.03480224363653196956 0.13880483535586193855 0.00000000000000000000 0.39479171208028279727
【思路】
由于给定范围可以视作无穷大，在所选的点的凸包的范围外的面积远大于凸包内，我们只需要考虑落在凸包上的点就好，因为在无穷大的区域里随机选点最近点是凸包内部点的概率无限趋近于0。因此先用Graham扫描法求凸包，注意保留落在边上的点，再求凸包各边指向外外部的法向量的夹角占2π的比例，最后按顺序输出即可。
【代码】
//************************************************************************ // File Name: F.cpp // Author: Shili_Xu // EMail: shili_xu@qq.com // Created Time: 2018年03月24日 星期六 22时32分59秒 //************************************************************************ #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const double EPS = 1e8, PI = acos(1); const int MAXN = 105; struct point { double x, y; int id; point() {} point(double _x, double _y, int _id = 1) : x(_x), y(_y), id(_id) {} bool operator<(const point &another) const; }; struct ans { double val; int id; ans() {} bool operator<(const ans &another) const { return id < another.id; } }; int tot, n; point p[MAXN], s[MAXN], normal[MAXN]; ans an[MAXN]; bool visited[MAXN]; int sgn(double x) { if (fabs(x) < EPS) return 0; return (x > 0 ? 1 : 1); } double det(const point &a1, const point &a2, const point &b1, const point &b2) { return ((a2.x  a1.x) * (b2.y  b1.y)  (a2.y  a1.y) * (b2.x  b1.x)); } double dot(const point &a1, const point &a2, const point &b1, const point &b2) { return ((a2.x  a1.x) * (b2.x  b1.x) + (a2.y  a1.y) * (b2.y  b1.y)); } double dist(const point &p1, const point &p2) { return sqrt((p2.x  p1.x) * (p2.x  p1.x) + (p2.y  p1.y) * (p2.y  p1.y)); } bool point::operator<(const point &another) const { if (sgn(det(p[0], *this, p[0], another)) == 0) return dist(p[0], *this) < dist(p[0], another); else return det(p[0], *this, p[0], another) > 0; } void graham() { double min_y = p[0].y, min_x = p[0].x; int index = 0; for (int i = 1; i <= n  1; i++) { if (p[i].y < min_y  (sgn(p[i].y  min_y) == 0 && p[i].x < min_x)) { min_y = p[i].y; min_x = p[i].x; index = i; } } swap(p[0], p[index]); sort(p + 1, p + n); s[0] = p[0]; s[1] = p[1]; tot = 2; for (int i = 2; i <= n  1; i++) { while (tot >= 2 && sgn(det(s[tot  1], p[i], s[tot  1], s[tot  2]) < 0)) tot; s[tot++] = p[i]; } } int main() { scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%lf %lf", &p[i].x, &p[i].y); p[i].id = i; } graham(); for (int i = 0; i < tot; i++) { if (i == 0) { normal[i].y = (s[i].x  s[tot  1].x); normal[i].x = s[i].y  s[tot  1].y; normal[i].id = s[i].id; } else { normal[i].y = (s[i].x  s[i  1].x); normal[i].x = s[i].y  s[i  1].y; normal[i].id = s[i].id; } } point zero(0, 0); for (int i = 0; i < tot; i++) { if (i == tot  1) { an[i].val = acos(dot(zero, normal[i], zero, normal[0]) / dist(zero, normal[i]) / dist(zero, normal[0])) / PI / 2; an[i].id = normal[i].id; } else { an[i].val = acos(dot(zero, normal[i], zero, normal[i + 1]) / dist(zero, normal[i]) / dist(zero, normal[i + 1])) / PI / 2; an[i].id = normal[i].id; } } memset(visited, false, sizeof(visited)); for (int i = 0; i < tot; i++) visited[s[i].id] = true; for (int i = 0; i < n; i++) if (!visited[i]) { an[tot].val = 0.0000000; an[tot++].id = i; } sort(an, an + n); for (int i = 0; i < n; i++) printf("%.8f\n", an[i].val); return 0; }

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