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Posted on 2009-09-20 00:53 Uriel 阅读(471) 评论(0) 编辑 收藏 引用 所属分类: POJ 、 计算几何
简单凸包。。凸包周长加半径L圆的周长。。
 /**//*Problem: 1113 User: Uriel
Memory: 636K Time: 47MS
Language: G++ Result: Accepted*/
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#include<math.h>
#include<stdio.h>
#include <stdlib.h>
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#define eps 1e-8
#define PI 3.141592653589793
#define zero(x) (((x)>0?(x):-(x))<eps)
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 struct point {
double x,y;
};
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point P[3001],convex[3001];
double res;
int N,L;
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//计算cross product (P1-P0)x(P2-P0)
 double xmult(point p1,point p2,point p0) {
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
//graham算法顺时针构造包含所有共线点的凸包,O(nlogn)
point p1,p2;
 int graham_cp(const void* a,const void* b) {
double ret=xmult(*((point*)a),*((point*)b),p1);
return zero(ret)?(xmult(*((point*)a),*((point*)b),p2)>0?1:-1):(ret>0?1:-1);
}
 void _graham(int n,point* p,int& s,point* ch) {
int i,k=0;
for (p1=p2=p[0],i=1;i<n;p2.x+=p[i].x,p2.y+=p[i].y,i++)
if (p1.y-p[i].y>eps||(zero(p1.y-p[i].y)&&p1.x>p[i].x))
p1=p[k=i];
p2.x/=n,p2.y/=n;
p[k]=p[0],p[0]=p1;
qsort(p+1,n-1,sizeof(point),graham_cp);
for (ch[0]=p[0],ch[1]=p[1],ch[2]=p[2],s=i=3;i<n;ch[s++]=p[i++])
for (;s>2&&xmult(ch[s-2],p[i],ch[s-1])<-eps;s--);
}
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//构造凸包接口函数,传入原始点集大小n,点集p(p原有顺序被打乱!)
//返回凸包大小,凸包的点在convex中
//参数maxsize为1包含共线点,为0不包含共线点,缺省为1
//参数clockwise为1顺时针构造,为0逆时针构造,缺省为1
//在输入仅有若干共线点时算法不稳定,可能有此类情况请另行处理!
//不能去掉点集中重合的点
 int graham(int n,point* p,point* convex,int maxsize=1,int dir=1) {
point* temp=new point[n];
int s,i;
_graham(n,p,s,temp);
for (convex[0]=temp[0],n=1,i=(dir?1:(s-1));dir?(i<s):i;i+=(dir?1:-1))
if (maxsize||!zero(xmult(temp[i-1],temp[i],temp[(i+1)%s])))
convex[n++]=temp[i];
delete []temp;
return n;
}
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double Dis(int a,int b)
  {
return sqrt((convex[a].y-convex[b].y)*(convex[a].y-convex[b].y)+(convex[a].x-convex[b].x)*(convex[a].x-convex[b].x));
}
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int main()
  {
scanf("%d %d",&N,&L);
for(int i=0;i<N;i++)
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scanf("%lf %lf",&P[i].x,&P[i].y);
}
int M=graham(N,P,convex,1,1);
res=0;
for(int i=0;i<M-1;i++)
 {
// printf("*%.2f %.2f*\n",convex[i].x,convex[i].y);
res+=Dis(i,i+1);
}
res+=Dis(0,M-1);
res+=2*PI*L;
printf("%.0f\n",res);
// system("PAUSE");
return 0;
}
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