66 lines
1.6 KiB
C++
66 lines
1.6 KiB
C++
#include <iostream>
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#include <cmath>
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#include <cstdio>
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#include <algorithm>
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#include<stack>
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using namespace std;
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const double pi=acos(-1.00),eps = 1e-6;//pi3.14~
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const int N=1001,inf=0x3f3f3f3f;
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double ans=0;
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struct dot{
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int x,y;
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}d[N];
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stack<int>s;
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int sign(double d)
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{
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if(fabs(d)<eps) return 0;
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return d>0? 1:-1;
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}
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double cross(double x1, double y1, double x2, double y2)
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{
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return x1 * y2 - x2 * y1;
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}
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double length(int x1,int y1,int x2,int y2)
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{
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int x=x1-x2,y=y1-y2;
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return sqrt((double)x*x+y*y);//CE double
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}
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int main()
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{
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int n,l,minn=1,f,ff,x0,x1,x2,y1,y2,y0;
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scanf("%d%d",&n,&l);
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for(int i=1;i<=n;i++)
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{
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scanf("%d%d",&d[i].x,&d[i].y);//&!!!
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minn=d[i].y<d[minn].y? i:minn;
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}
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int j=(n+minn-2)%n+1;
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// s.push(minn%n+1);
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s.push(minn);
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s.push(j);
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ans+=length(d[j].x,d[j].y,d[minn].x,d[minn].y);//漏了初始的边
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while(j!=minn)
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{
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j=(j+n-2)%n+1;
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f=s.top();
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s.pop();
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ff=s.top();
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while(0>=sign(cross(d[f].x-d[ff].x,d[f].y-d[ff].y,d[j].x-d[ff].x,d[j].y-d[ff].y)))
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{
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ans-=length(d[f].x,d[f].y,d[ff].x,d[ff].y);
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// printf("--%d(%d,%d),%d(%d,%d): -%lf=%lf\n",f,d[f].x,d[f].y,ff,d[ff].x,d[ff].y,length(d[f].x,d[f].y,d[ff].x,d[ff].y),ans);
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f=s.top();
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s.pop();
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if(s.empty()) break;//RE
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ff=s.top();
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}
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ans+=length(d[j].x,d[j].y,d[f].x,d[f].y);
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// printf("++%d(%d,%d),%d(%d,%d): +%lf=%lf\n",j,d[j].x,d[j].y,f,d[f].x,d[f].y,length(d[j].x,d[j].y,d[f].x,d[f].y),ans);
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s.push(f);
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s.push(j);
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}
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// printf("%d ",s.size());
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printf("%.0lf",ans+2*l*pi);//WA*2 （周长跟多边形形状有关）错了，无关，自带四舍五入
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return 0;
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}
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