poj1584

题意：已知一个多边形的n个顶点坐标，然后再给一个钉子，给定钉子的半径和圆心坐标，首先判断多边形是否为凸多边形，若为凸多边形，再判断钉子是否可以放到凸多边形内部。
 
思路：
1.由于顶点给出的顺序可能为逆时针或顺时针，所以在判断是否为凸边形，利用叉积的性质，每个点依次扫描看是否都顺着同一的方向转动。
2.判断圆(钉子)是否在多边形内部，第一步判断圆心是否在凸多边形内部（以圆心为定点扫描一周），第二步再判断圆心到某一边的最短距离，若存在某最短距离大于圆心，则圆不能放在凸多边内。

或者思路可以参考这个：

http://blog.csdn.net/lyy289065406/article/details/6648606

#include <stdlib.h>
#include <math.h>
#include <iostream>

#define MAXN 1000
#define offset 10000
#define eps 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
#define _sign(x) ((x)>eps?1:((x)<-eps?2:0))


struct point{ double x, y; };
struct line{ point a, b; };

double xmult(point p1, point p2, point p0){
	return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
}

//判定凸多边形,顶点按顺时针或逆时针给出,允许相邻边共线
bool is_convex(int n, point* p){
	int i, s[3] = { 1, 1, 1 };
	for (i = 0; i < n && s[1] | s[2]; i++)
		s[_sign(xmult(p[(i + 1)%n], p[(i + 2)%n], p[i]))] = 0;
	return s[1] | s[2];
}

//判定凸多边形,顶点按顺时针或逆时针给出,不允许相邻边共线
bool is_convex_v2(int n, point* p){
	int i, s[3] = { 1, 1, 1 };
	for (i = 0; i < n && s[0] && s[1] | s[2]; i++)
		s[_sign(xmult(p[(i + 1)%n], p[(i + 2)%n], p[i]))] = 0;
	return s[0] && s[1] | s[2];
}

//判点在凸多边形内或多边形边上,顶点按顺时针或逆时针给出
bool inside_convex(point q, int n, point* p){
	int i, s[3] = { 1, 1, 1 };
	for (i = 0; i < n && s[1] | s[2]; i++)
		s[_sign(xmult(p[(i + 1)%n], q, p[i]))] = 0;
	return s[1] | s[2];
}

//判点在凸多边形内,顶点按顺时针或逆时针给出,在多边形边上返回0
bool inside_convex_v2(point q, int n, point* p){
	int i, s[3] = { 1, 1, 1 };
	for (i = 0; i < n && s[0] && s[1] | s[2]; i++)
		s[_sign(xmult(p[(i + 1)%n], q, p[i]))] = 0;
	return s[0] && s[1] | s[2];
}

//判点在任意多边形内,顶点按顺时针或逆时针给出
//on_edge表示点在多边形边上时的返回值,offset为多边形坐标上限
bool inside_polygon(point q, int n, point* p, int on_edge = 1){
	point q2;
	int i = 0, count;
	while (i < n)
		for (count = i = 0, q2.x = rand() + offset, q2.y = rand() + offset; i < n; i++)
			if (zero(xmult(q, p[i], p[(i + 1)%n])) && (p[i].x - q.x)*(p[(i + 1)%n].x - q.x) < eps && (p[i].y - q.y)*(p[(i + 1)%n].y - q.y) < eps)
				return on_edge;
			else if (zero(xmult(q, q2, p[i])))
				break;
			else if (xmult(q, p[i], q2)*xmult(q, p[(i + 1)%n], q2) < -eps && xmult(p[i], q, p[(i + 1)%n])*xmult(p[i], q2, p[(i + 1)%n]) < -eps)
				count++;
			return count & 1;
}

inline bool opposite_side(point p1, point p2, point l1, point l2){
	return xmult(l1, p1, l2)*xmult(l1, p2, l2) < -eps;
}

inline bool dot_online_in(point p, point l1, point l2){
	return zero(xmult(p, l1, l2)) && (l1.x - p.x)*(l2.x - p.x) < eps && (l1.y - p.y)*(l2.y - p.y) < eps;
}

//判线段在任意多边形内,顶点按顺时针或逆时针给出,与边界相交返回1
bool inside_polygon(point l1, point l2, int n, point* p){
	point t[MAXN], tt;
	int i, j, k = 0;
	if (!inside_polygon(l1, n, p) || !inside_polygon(l2, n, p))
		return 0;
	for (i = 0; i < n; i++)
		if (opposite_side(l1, l2, p[i], p[(i + 1)%n]) && opposite_side(p[i], p[(i + 1)%n], l1, l2))
			return 0;
		else if (dot_online_in(l1, p[i], p[(i + 1)%n]))
			t[k++] = l1;
		else if (dot_online_in(l2, p[i], p[(i + 1)%n]))
			t[k++] = l2;
		else if (dot_online_in(p[i], l1, l2))
			t[k++] = p[i];
		for (i = 0; i < k; i++)
			for (j = i + 1; j < k; j++){
				tt.x = (t[i].x + t[j].x) / 2;
				tt.y = (t[i].y + t[j].y) / 2;
				if (!inside_polygon(tt, n, p))
					return 0;
			}
			return 1;
}

double distance(point p1, point p2)
{
	return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));
}
double disptoline(point p, point l1, point l2){
	return fabs(xmult(p, l1, l2)) / distance(l1, l2);
}
int intersect_seg_circle(point c, double r, point l1, point l2){
	double t1 = distance(c, l1) - r, t2 = distance(c, l2) - r;
	point t = c;
	if (t1<eps || t2<eps)
		return t1>-eps || t2>-eps;
	t.x += l1.y - l2.y;
	t.y += l2.x - l1.x;
	return xmult(l1, c, t)*xmult(l2, c, t) < eps && disptoline(c, l1, l2) - r < eps;
}
//判断圆是否在多边形内
bool circle_in_polygen(double r, point o, int n, point *p)
{
	for (int i = 1; i < n; i++)
	{
		bool flag = intersect_seg_circle(o, r, p[i - 1], p[i]);
		if (flag) { return false; break; }
	}
	return true;
}

point intersection(line u, line v){
	point ret = u.a;
	double t = ((u.a.x - v.a.x)*(v.a.y - v.b.y) - (u.a.y - v.a.y)*(v.a.x - v.b.x))
		/ ((u.a.x - u.b.x)*(v.a.y - v.b.y) - (u.a.y - u.b.y)*(v.a.x - v.b.x));
	ret.x += (u.b.x - u.a.x)*t;
	ret.y += (u.b.y - u.a.y)*t;
	return ret;
}

point barycenter(point a, point b, point c){
	line u, v;
	u.a.x = (a.x + b.x) / 2;
	u.a.y = (a.y + b.y) / 2;
	u.b = c;
	v.a.x = (a.x + c.x) / 2;
	v.a.y = (a.y + c.y) / 2;
	v.b = b;
	return intersection(u, v);
}

//多边形重心
point barycenter(int n, point* p){
	point ret, t;
	double t1 = 0, t2;
	int i;
	ret.x = ret.y = 0;
	for (i = 1; i<n - 1; i++)
		if (fabs(t2 = xmult(p[0], p[i], p[i + 1]))>eps){
			t = barycenter(p[0], p[i], p[i + 1]);
			ret.x += t.x*t2;
			ret.y += t.y*t2;
			t1 += t2;
		}
		if (fabs(t1) > eps)
			ret.x /= t1, ret.y /= t1;
		return ret;
}

int main()
{
	int n;
	double pegR;
	point peg;
	while (std::cin >> n && (n >= 3))
	{
		std::cin >> pegR >> peg.x >> peg.y;
		point p[155];
		for (int i = 0; i < n; i++)
		{
			std::cin >> p[i].x >> p[i].y;
		}
		bool flag = is_convex(n, p);
		bool flag2 = inside_convex_v2(peg, n, p);
		bool flag3 = circle_in_polygen(pegR, peg, n, p);

		if (!flag)
			std::cout << "HOLE IS ILL-FORMED" << std::endl;
		else
		{
			if (flag2 && flag3)
				std::cout << "PEG WILL FIT" << std::endl;
			else
				std::cout << "PEG WILL NOT FIT" << std::endl;
		}
	}
}