poj 1584 A Round Peg in a Ground Hole 点到直线的距离 点是否在多边形内 多边形是否为凸

http://poj.org/problem?id=1584

给一个多边形，和一个圆形的钉子，判断多边形是否为凸多边形，若为凸多边形判断钉子能否被包含在多边形内。

判断是否为凸多边形：相邻的两条边做叉乘i*(i+1)都大于零（假定顶点为逆时针排序，顺时针反之）则为凸多边形。

判断圆心是否在多边形内：设o为圆心，s1,s2为相邻顶点，若叉乘（s1s2）*(s1o)都大于零则圆心在多边形内部。（假定顶点为逆时针排序，顺时针反之）

之后找圆心到多边形边的最短距离，若最短距离小于圆半径必定不会包含在多边形内。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define eps 1e-8
using namespace std;
struct point
{
    double x,y;
}p[10000];
double multi(point p0,point p1,point p2)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
point inter(point u1,point u2,point v1,point v2)
{
    point ret=u1;
    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
        /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
    ret.x+=(u2.x-u1.x)*t;
    ret.y+=(u2.y-u1.y)*t;
    return ret;
}

bool is(point s1,point e1,point s2,point e2)
{
    return  (max(s1.x,e1.x)>=min(s2.x,e2.x))&&
        (max(s2.x,e2.x)>=min(s1.x,e1.x))&&
        (max(s1.y,e1.y)>=min(s2.y,e2.y))&&
        (max(s2.y,e2.y)>=min(s1.y,e1.y))&&
        (multi(s1,s2,e1)*multi(s1,e1,e2)>=0)&&
        (multi(s2,s1,e2)*multi(s2,e2,e1)>=0);
}
point pt(point p,point t1,point t2)
{
    point t=p;
    t.x+=t1.y-t2.y;
    t.y+=t2.x-t1.x;
    return inter(p,t,t1,t2);
}
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void work(point t,int n,double r)
{
    int i;
    double flag;
    point a;
    flag=multi(p[0],p[1],p[2]);//做顶点顺、逆时针标记
    for(i=1;i<n-1;i++)
        if(flag*multi(p[i],p[i+1],p[i+2])<-eps)
        {
            printf("HOLE IS ILL-FORMED\n");
            return;
        } 
        for(i=0;i<n;i++)
        {
            a=pt(t,p[i],p[i+1]);
            if(multi(p[i],p[i+1],t)*flag<-eps)//圆心不在多边形内部
            {
                printf("PEG WILL NOT FIT\n");
                return;
            }
            if(dis(a,t)-r<-eps)
            {
                printf("PEG WILL NOT FIT\n");
                return;
            }
        }
        printf("PEG WILL FIT\n");
}
int main()
{
    int i,n;
    double r;
    point t;
    while(scanf("%d",&n),n>=3)
    {
        scanf("%lf%lf%lf",&r,&t.x,&t.y);
        for(i=0;i<n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        p[n]=p[0];
        work(t,n,r);
    }
    return 0;
}