UVA 218 Moth Eradication(凸包应用)

UVA 218 Moth Eradication(凸包应用)"/>

UVA 218 Moth Eradication(凸包应用)

UVA 218 Moth Eradication(凸包应用)

题意:

       给你n个点的集合,要你求出这个点集的凸包(求凸包最小点集),并且按时针输出所有点,且输出该凸包的周长.

分析:

       直接用刘汝佳训练指南P272求凸包的模板即可,不过注意要小心n=1或2的情况. 刘汝佳的模板可以直接处理n=>1的所有情况.

       注意:本题的输出点从x坐标最小的值开始(x坐标相同就输出y坐标最小的)按顺时针顺序输出所有点坐标. 两个输出实例之间要有空格.

AC代码:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e-10;
int dcmp(double x)
{if(fabs(x)<eps) return 0;return x<0?-1:1;
}
struct Point
{double x,y;Point(){}Point(double x,double y):x(x),y(y){}bool operator==(const Point& rhs)const{return dcmp(x-rhs.x)==0 && dcmp(y-rhs.y)==0;}bool operator<(const Point& rhs)const{return dcmp(x-rhs.x)<0 || (dcmp(x-rhs.x)==0 && dcmp(y-rhs.y)<0);}
};
typedef Point Vector;
double dist(Point a,Point b)
{return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
Vector operator-(Point A,Point B)
{return Vector(A.x-B.x,A.y-B.y);
}
double Cross(Vector A,Vector B)
{return A.x*B.y-A.y*B.x;
}
int ConvexHull(Point *p,int n,Point *ch)
{sort(p,p+n);n=unique(p,p+n)-p;int m=0;for(int i=0;i<n;i++){while(m>1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-1])<=0) m--;ch[m++]=p[i];}int k=m;for(int i=n-2;i>=0;i--){while(m>k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-1])<=0) m--;ch[m++]=p[i];}if(n>1) m--;return m;
}const int maxn=10000+5;
Point p[maxn],ch[maxn];int main()
{int n,kase=0;while(scanf("%d",&n)==1 && n){if(kase>0) printf("\n");for(int i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);int m=ConvexHull(p,n,ch);double ans=0;printf("Region #%d:\n",++kase);printf("(%.1lf,%.1lf)",ch[0].x,ch[0].y);for(int i=m-1;i>=0;i--){printf("-(%.1lf,%.1lf)",ch[i].x,ch[i].y);ans += dist(ch[i],ch[(m+i-1)%m]);}printf("\n");if(m==1) ans=0;else if(m==2) ans/=2;printf("Perimeter length = %.2lf\n",ans);}return 0;
}