A Round Peg in a Ground Hole(圆与凸包)

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

题意：判断所给的点能不能形成凸包，并判断所给的圆是否在凸包内。

改了好几天的一个题，今天才发现是输入顺序弄错了，办过的最脑残的事情。。sad

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;
const int N=1002;
const double eps=1e-8;
double pi=acos(-1.0);
int n;
struct point
{
    double x,y;
    point(double x = 0,double y = 0):x(x),y(y) {}
    double norm()//向量的模
    {
        return sqrt(x*x+y*y);
    }

};
point operator-(const point &a,const point &b)
{
    return point(a.x-b.x,a.y-b.y);
}
int cmp(double x)//精度处理
{
    if (fabs(x) < eps)
        return 0;
    if (x > 0)
        return 1;
    return -1;

}
double det(const point &a,const point &b)//叉乘
{
    return a.x*b.y-a.y*b.x;
}

double dot(const point &a,const point &b)//点乘
{
    return a.x*b.x+a.y*b.y;
}

double dist(const point &a,const point &b)//两点间的距离
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool PointOnSegment(point p,point s,point t)//判断点是否在线段上
{
    return cmp(det(p-s,t-s))==0&&cmp(dot(p-s,p-t))<=0;
}

double dis_point_segment(const point p,const point s,const point t)//点到线段的距离
{
    if(cmp(dot(p-s,t-s))<0) return (p-s).norm();
    if(cmp(dot(p-t,s-t))<0) return (p-t).norm();
    return fabs(det(s-p,t-p)/dist(s,t));
}
bool is_convex(point *p)//判断凸包
{
    int pre = 0;
    p[n] = p[0];
    p[n+1] = p[1];
    for (int i = 2; i <= n; i++)
    {
        int dir = cmp(det(p[i-1]-p[i-2],p[i]-p[i-1]));
        if (!pre)
            pre = dir;
        if (pre*dir < 0) return false;
    }
    return true;
}
int point_in(point t,point *ch)//判断点是否在凸包内
{
    int num=0,d1,d2,k;
    ch[n]=ch[0];
    for(int i=0; i<n; i++)
    {
        if(PointOnSegment(t,ch[i],ch[i+1])) return 2;
        k=cmp(det(ch[i+1]-ch[i],t-ch[i]));
        d1=cmp(ch[i].y-t.y);
        d2=cmp(ch[i+1].y-t.y);
        if(k>0&&d1<=0&&d2>0) num++;
        if(k<0&&d2<=0&&d1>0) num--;
    }
    return num!=0;
}
int main()
{
    double x,y,r;
    while(~scanf("%d",&n))
    {
        if (n < 3)
            break;
        point p[N];
        scanf("%lf%lf%lf",&r,&x,&y);
        point t(x,y);
        for (int i = 0; i < n; i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        p[n] = p[0];//连接首尾的点
        if (!is_convex(p))
        {
            printf("HOLE IS ILL-FORMED
");
            continue;
        }
        if(point_in(t,p))
        {
            double Min = dis_point_segment(t,p[0],p[1]);
            for (int i = 1; i < n; i++)
            {
                double dis = dis_point_segment(t,p[i],p[i+1]);
                Min = min(dis,Min);//圆心到所有线段的最小距离
            }
            if (cmp(Min-r)>= 0)
                printf("PEG WILL FIT
");
            else
                printf("PEG WILL NOT FIT
");
        }
        else
            printf("PEG WILL NOT FIT
");

    }
    return 0;
}