poj 1584 A Round Peg in a Ground Hole

来源：互联网发布：手机淘宝尺码助手编辑：程序博客网时间：2024/06/05 23:03

题意：按照顺时针或者逆时针给你一个点集首先要判断是否为凸包，如果为凸包的话，就判断圆能否完全放在这个凸包内

思路：凸包的话叉积跑一边就ok，注意在输入点是记得封口；

判断圆是否在凸包内，首先判断圆心是否在凸包内，然后就简单了；

要注意的一点是先输入半径，再输入的坐标。。。。哥白贡献了一个WA

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define inf 0x7ffffffusing namespace std;const double eps=1e-10;const double pi=acos(-1.0);struct P{    double x,y;int index;    P () {}    P (double x,double y):x(x),y(y){    }    P operator -(P p){        return P(x-p.x,y-p.y);    }    P operator +(P p){        return P(x+p.x,y+p.y);    }    P operator *(double d){        return P(x*d,y*d);    }    double det(P p){        return x*p.y-y*p.x;    }};P p[1000000],r;double R;int n;double dis;double dist(P a,P b,P c){   double A,B,C;    B=1.0;    A=-((b.y-a.y)/(b.x-a.x));    C=-(A*a.x+a.y);if(b.x-a.x==0){A=1.0;B=0.0;C=-a.x;}    double mm=fabs(A*c.x+B*c.y+C);    double aaa=sqrt(A*A+B*B);//printf("\n");//printf("A== %lf B== %lf C== %lf\n",A,B,C);    return mm/aaa;}bool judge1(){    int k=0;    P q;    q.y=r.y;    q.x=-10000000.0;    for(int i=0;i<n-1;i++)    {//printf("%lf %lf\n",p[i].x,p[i].y);//printf("%lf %lf\n",p[i+1].x,p[i+1].y);        if(p[i].y==p[i+1].y)        continue;        if(p[i].y<p[i+1].y)        {            if(p[i].y==r.y)            continue;        }        else        {            if(p[i+1].y==r.y)            continue;        }        if((q-p[i]).det(p[i+1]-p[i])*(r-p[i]).det(p[i+1]-p[i])<=0 && (p[i]-q).det(r-q)*(p[i+1]-r).det(r-q)<=0)        k++;    }    if(k%2==0)    return 0;    return 1;}bool judge(){    if(!judge1())    return 0;    for(int i=0;i<n-1;i++)    {        if(dist(p[i],p[i+1],r)>=R)        continue;        else        return 0;    }    return 1;}int main(){    while(~scanf("%d",&n) && n>=3)    {        int fg=1,fg1=1;        scanf("%lf%lf%lf",&R,&r.x,&r.y);        for(int i=0;i<n;i++)        scanf("%lf%lf",&p[i].x,&p[i].y);        p[n]=p[0];        n++;        dis=(p[1]-p[0]).det(p[2]-p[0]);        for(int i=2;i<n;i++)        {            if(dis*((p[i-1]-p[i-2]).det(p[i]-p[i-2]))<0)            fg=0;        }        if(!fg)        printf("HOLE IS ILL-FORMED\n");        else        {            if(judge())            printf("PEG WILL FIT\n");            else            printf("PEG WILL NOT FIT\n");        }    }    return 0;}

0 0