POJ 1584 A Round Peg in a Ground Hole （凸包的判断+点在凸包+圆在凸包内）

大意：给定多边形的点和圆的半径长，和原点。进行凸包的判断+点在凸包+圆在凸包内的判断。

#include<map>#include<cmath>#include<queue>#include<cmath>#include<cstdio>#include<stack>#include<iostream>#include<cstring>#include<algorithm>#define inf 999999999.9#define eps 1e-8#define ls l,mid,rt<<1#define rs mid+1,rt,rt<<1|1#define LL __int64#define _sign(x) ((x)>eps?1:( (x)<-eps?2:0 )  )using namespace std;struct point{    double x,y;};struct Circle{    double r;    point poi;}c;struct Line{    double a,b,c;};double xmult(point p1,point p2,point p){    return (p1.x - p.x)*(p2.y-p.y) - (p2.x - p.x)*(p1.y - p.y);}int checkvex(int n,point *p){//多边形的判断    int s[3]={1,1,1};    for(int 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];}bool insideConvex(point q,int n,point *p){//点在多边形内的判断    int s[3] ={1,1,1};    for(int 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];}Line towpoLine(point p1,point p2){ //两点连城的线    Line l;    l.a = p1.y - p2.y;    l.b = p2.x - p1.x;    l.c = p1.x*p2.y - p2.x*p1.y;    return l;}double potoLine(point p,Line l){//点到直线的距离    return fabs(l.a*p.x+l.b*p.y+l.c)/sqrt(l.a*l.a+l.b*l.b);}point q[10010];int main(){    int n,i,j,k;    while(~scanf("%d",&n),n>2){        scanf("%lf%lf%lf",&c.r,&c.poi.x,&c.poi.y);        for(i = 0;i < n;++ i)            scanf("%lf%lf",&q[i].x,&q[i].y);        if(checkvex(n,q)){//判断在多边形上            if(insideConvex(c.poi,n,q)){//圆心在多边形内                for(i = 0;i < n;++ i){                    Line l = towpoLine(q[i],q[(i+1)%n]);                    if(potoLine(c.poi,l) - c.r <0 ){//不知道为什么< eps就错                        printf("PEG WILL NOT FIT\n");                        break;                    }                }                if(i == n)                    printf("PEG WILL FIT\n");            }            else                printf("PEG WILL NOT FIT\n");        }        else            printf("HOLE IS ILL-FORMED\n");    }    return 0;}