poj 1113 Wall

#include <iostream>
#include <algorithm>
#include <cmath>
#define PI 3.1415926
using namespace std;
struct point 
{
    int x,y;
}st;
point castle[1005];
double dis(point a,point b)
{
    return sqrt(pow(a.x-b.x+0.0,2.0)+pow(a.y-b.y+0.0,2.0));
}
int cmp(const point& a,const point& b)
{
    if((b.x-st.x)*(a.y-st.y)==(a.x-st.x)*(b.y-st.y))
        return dis(st,a)<dis(st,b);
    else
        return (b.x-st.x)*(a.y-st.y)<(a.x-st.x)*(b.y-st.y);
}
bool left(point a,point b,point c)    
//a,b,c分别为 castle[s[t-1]],castle[s[t]],castle[i],
//如果直线ac的斜率大于ab的斜率,则说明从ab到ac做左转,即ac在ab的逆时针方向
{
    return (c.y-a.y)*(b.x-a.x) >= (b.y-a.y)*(c.x-a.x);    //(c.y-a.y)/(c.x-a.x) > (b.y-a.y)/(b.x-a.x)
    //要特别注意=的情况，相等说明c在直线ab上,判断左转时这种同直线的情况也要考虑,若只是 > ，则 RE
}
int main()
{
    int n,l,idx=0;
    cin>>n>>l;
    for(int i=0;i<n;++i)
    {
        cin>>castle[i].x>>castle[i].y;
        if(castle[i].y<castle[idx].y||(castle[i].y==castle[idx].y&&castle[i].x<castle[idx].x))
            idx=i;
    }
    st.x=castle[idx].x;st.y=castle[idx].y;
    sort(castle,castle+n,cmp);
    int t=-1;
    int s[1005];
    s[++t]=0;s[++t]=1;s[++t]=2;
    for(int i=3;i<n;++i)
    {
        while(!left(castle[s[t-1]],castle[s[t]],castle[i]))
            --t;
        s[++t]=i;
    }
    double len=0;
    for(int i=1;i<=t;++i)
        len+=dis(castle[s[i-1]],castle[s[i]]);
    len+=dis(castle[s[0]],castle[s[t]])+2*l*PI;
    printf("%.0f\n",len);
    return 0;
}