POJ1113(凸包)

题目链接: http://poj.org/problem?id=1113

 

题意:给出一多边形,求出能包围这个多边形并且距第点至少L距离的图形的周长.

 

题目分析:凸包题,其中,距内部多边形至少L可根据圆形的性质,相当于多加了

             一个r=L的圆的整个弧分开来连接凸包上所有边.

 

8494335dooder_daodao1113Accepted408K32MSG++1350B2011-04-15 20:24:08

 

 

#include<stdio.h>#include<stdlib.h>#include<math.h>#define M 1008const double eps = 1e-8;const double inf = 10e8;const double Pi = acos(-1.0);typedef struct{double x,y;}Point;Point p[M];int stk[M];int n,top;double l;int Sign(double x){if(fabs(x)<eps) return 0;if(x>0) return 1;return -1;}double Dis2(Point p,Point q){return (p.x-q.x)*(p.x-q.x)+(p.y-q.y)*(p.y-q.y);}double Multi(Point p,Point q,Point o){return (p.x-o.x)*(q.y-o.y)-(q.x-o.x)*(p.y-o.y);}int Cmp(const void *a,const void *b){Point *x=(Point *)a;Point *y=(Point *)b;double d=Multi(*x,*y,p[0]);if(Sign(d)==0)return Dis2(*x,p[0])>Dis2(*y,p[0])?1:-1;return d>0?-1:1;}int main(){Point tmp;int i,k;double ans;while(scanf("%d%lf",&n,&l)!=EOF){for(i=0,k=0;i<n;i++){scanf("%lf%lf",&p[i].x,&p[i].y);if(Sign(p[i].y-p[k].y)<0||Sign(p[i].y-p[k].y)==0&&Sign(p[i].x-p[k].x)<0)k=i;}tmp=p[k],p[k]=p[0],p[0]=tmp;qsort(p+1,n-1,sizeof(p[0]),Cmp);for(top=0;top<2;stk[top]=top++);for(i=2;i<n;i++){while(Multi(p[stk[top-2]],p[stk[top-1]],p[i])<eps&&top>1)top--;stk[top++]=i;}for(i=0,ans=0.0;i<top;i++)ans+=sqrt(Dis2(p[stk[i]],p[stk[(i+1)%top]]));ans+=Pi*l*2;printf("%.0f/n",ans);}return 0;}