POJ 1584

/*************************************************************************
    > File Name:            poj_1584_back.cpp
    > Author:               Howe_Young
    > Mail:                 1013410795@qq.com
    > Created Time:         2015年04月06日 星期一 19时03分03秒
 ************************************************************************/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <cstdio>
#define EPS 1e-8
using namespace std;
struct point{
    double x, y;
};
const int N = 1000;
point p[N];
int sgn(double d)
{
    if (fabs(d) < EPS)
        return 0;
    return d > 0 ? 1 : -1;
}
double x_mutli(point p1, point p2)
{
    return (p1.x * p2.y - p2.x * p1.y);
}
double dot_multi(point p1, point p2)
{
    return (p1.x * p2.x + p1.y * p2.y);
}
//方向
double get_direction(point p1, point p2, point p3)
{
    point v1, v2;
    v1.x = p3.x - p1.x; v1.y = p3.y - p1.y;
    v2.x = p2.x - p1.x; v2.y = p2.y - p1.y;
    return x_mutli(v1, v2);
}
//得到模长
double get_length_of_mold(point p)
{
    return sqrt(p.x * p.x + p.y * p.y);
}
bool is_vonvex(int n)
{
    double tmp1 = 0.0, tmp2;
    for (int i = 0; i < n; i++)//因为凸多边形的相邻边的拐向都相同,要么都顺时针,要么多逆时针
    {
        tmp2 = sgn(get_direction(p[i], p[(i+1)%n], p[(i+2)%n]));
        if (tmp1 * tmp2 < -EPS)
            return false;
        tmp1 = tmp2;
    }
    return true;
}
//叉积判断点是否在多边形内,只适合凸多边形
bool point_in_polygon(point peg, int n)
{
    double tmp1 = 0.0, tmp2;
    for (int i = 0; i < n; i++)
    {
        tmp2 = sgn(get_direction(p[i], p[(i+1)%n], peg));
        if (tmp1 * tmp2 < -EPS)
            return false;
        tmp1 = tmp2;
    }
    return true;
}
//判断圆是否在多边形内,就是判断点到边的最小离跟半径的关系
bool circle_in_polygon(point peg, double peg_r, int n)
{
    if (peg_r == 0.0)
        return true;
    double shadow;//v1向量在v2向量上的投影长度 a点乘b然后除以b的模就是a在b上的投影
    point v1, v2;
    double ans;
    for (int i = 0; i < n; i++)
    {
        v1.x = peg.x - p[i].x; v1.y = peg.y - p[i].y;
        v2.x = p[(i+1)%n].x - p[i].x; v2.y = p[(i+1)%n].y - p[i].y;
        shadow = dot_multi(v1, v2) / (v2.x * v2.x + v2.y * v2.y) * 1.0;//这里是求投影占v2向量模的长度 的比例,如果大于1,或者小于0, 垂足肯定在外面了
        if (shadow >= 0.0 && shadow <= 1.0)//利用面积来求高,也就是距离,叉乘的绝对值是三角形面积的两倍
            ans = fabs(x_mutli(v1, v2)) / get_length_of_mold(v2);
        else
        {
            //如果垂足在外面,找最近的一个端点
            v2.x = peg.x - p[(i+1)%n].x; v2.y = peg.y - p[(i+1)%n].y;
            ans = sgn(get_length_of_mold(v1) - get_length_of_mold(v2)) == -1 ? get_length_of_mold(v1) : get_length_of_mold(v2);
        }
        if (ans - peg_r < -EPS)//如果相交,返回false
            return false;
    }
    return true;

}
int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int n;
    point peg;
    double peg_r;
    while (~scanf("%d", &n) && n >= 3)
    {
        scanf("%lf %lf %lf", &peg_r, &peg.x, &peg.y);
        for (int i = 0; i < n; i++)
            scanf("%lf %lf", &p[i].x, &p[i].y);
        if (!is_vonvex(n))//判断是否是凸多边形
        {
            puts("HOLE IS ILL-FORMED");
            continue;
        }
        if (point_in_polygon(peg , n) && circle_in_polygon(peg, peg_r, n))
        {
            puts("PEG WILL FIT");
        }
        else
            puts("PEG WILL NOT FIT");
    }
    return 0;
}