题目地址:POJ 1113
先求出凸包的周长,然后剩下的弧合起来一定是个半径为l的圆,然后再加上以l为半径的圆的周长即可。
代码如下:
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
using namespace std;
#define LL long long
const int INF=0x3f3f3f3f;
int n, top, l;
double PI=acos(-1.0);
struct Point
{
int x, y;
}p[1010], tu[1010];
double dist(Point x, Point y)
{
return sqrt((x.x-y.x)*(x.x-y.x)*1.0+(x.y-y.y)*(x.y-y.y));
}
Point operator - (Point x, Point y)
{
Point z;
z.x=x.x-y.x;
z.y=x.y-y.y;
return z;
}
int Cross(Point x, Point y)
{
return x.x*y.y-x.y*y.x;
}
int cmp(Point x, Point y)
{
if(x.x==y.x) return x.y<y.y;
return x.x<y.x;
}
void Andew()
{
int i, j, k;
sort(p,p+n,cmp);
top=0;
for(i=0;i<n;i++)
{
while(top>1&&Cross(tu[top-1]-tu[top-2],p[i]-tu[top-1])<0) top--;
tu[top++]=p[i];
}
k=top;
for(i=n-2;i>=0;i--)
{
while(top>k&&Cross(tu[top-1]-tu[top-2],p[i]-tu[top-1])<0) top--;
tu[top++]=p[i];
}
double ans=0;
for(i=1;i<top;i++)
{
ans+=dist(tu[i],tu[i-1]);
//printf("%.2f\n",ans);
}
ans+=2*PI*l;
printf("%d\n",(int)(ans+0.5));
}
int main()
{
int i, j;
scanf("%d%d",&n,&l);
for(i=0;i<n;i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
}
Andew();
return 0;
}