POJ 1584[A Round Peg in a Ground Hole]题解
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题目梗概
题目给出了一个多边形,和一个已知半径和圆心的圆,判断:
1.该多边形是否是凸多边形。
2.这个圆能否被这个多边形完全覆盖。
(传送门)
那么需要解决的,有3个问题。
1.该多边形是否是凸多边形。
2.圆心(某个点)是否在凸多边形内
3.圆心(某个点)对所有线段的距离是否都小于圆的半径。
关于问题1:在学习笔记2中有提过,打个广告;
关于问题2,本来是有通解的,但是既然验证时凸多边形通过滴,只需利用叉积判断是否在所有边的同一侧即可。
关于问题3:在学习笔记1中有提过,打个广告;
搞定
哦还有,打个广告,不,膜拜—— Orz ZH大佬 Orz Lynstery
代码
#include<cstdio>#include<cmath>using namespace std;int n;double R,eps=1e-10;struct data{ double x,y; data(double x=0,double y=0):x(x),y(y) { }}o,a[10005];double absf(double x){return (x>0)?x:-x;}int fcmp(double x,double y) {if (absf(x-y)<eps) return 0; return (x<y)?-1:1;}data operator + (const data a,const data b){return data(a.x+b.x,a.y+b.y);}data operator - (const data a,const data b){return data(a.x-b.x,a.y-b.y);}data operator * (const data a,double b){return data(a.x*b,a.y*b);}double dot(const data a,const data b) {return a.x*b.x+a.y*b.y;}double leng(const data a){return sqrt(dot(a,a));}double cross(const data a,const data b) {return a.x*b.y-a.y*b.x;}double dislne(data p,data a,data b){ data Va=b-a,Vb=p-a; return absf(cross(Va,Vb))/leng(Va);}bool check_Tu(){ bool f=true; for (int i=1;i<=n;i++) if (fcmp(cross(a[i%n+1]-a[i],a[(i-2+n)%n+1]-a[i]),0)<0) {f=false; break;} if (f) return true; for (int i=1;i<=n;i++) if (fcmp(cross(a[(i+n-2)%n+1]-a[i],a[i%n+1]-a[i]),0)<0) return false; return true;}bool _check(){ bool ga,gb; ga=gb=true; for (int i=1;i<=n;i++) if (fcmp(cross(o-a[i],o-a[i%n+1]),0)<0) {ga=false; break;} for (int i=1;i<=n;i++) if (fcmp(cross(o-a[i%n+1],o-a[i]),0)<0) {gb=false; break;} if (!ga&&!gb) return false; for (int i=1;i<=n;i++) if (fcmp(dislne(o,a[i],a[i%n+1]),R)<0) return false; return true;}int main(){ double x,y; for (scanf("%d",&n);n>2;scanf("%d",&n)) { scanf("%lf%lf%lf",&R,&x,&y); o=data(x,y); for (int i=1;i<=n;i++){scanf("%lf%lf",&x,&y); a[i]=data(x,y);} if (!check_Tu()) printf("HOLE IS ILL-FORMED\n"); else if (!_check()) printf("PEG WILL NOT FIT\n"); else printf("PEG WILL FIT\n"); } return 0;}
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