hdu_1348_Wall（凸包）

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1348

题意：让你求n个点的凸包，凸包离点的距离为l

题解：就凸包周长+一个半径为l的圆周长

 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #define F(i,a,b) for(int i=a;i<=b;++i)
 using namespace std;
 /*
 * 求凸包，Graham算法  * 点的编号0~n-1
 * 返回凸包结果Stack[0~top-1]为凸包的编号
  */
 const int MAXN = ;
 const double eps = 1e-;
 const double PI = acos(-1.0);
 struct Point {
      double x,y;
      Point(){}
      Point(double _x,double _y){x = _x,y = _y;}
      Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}
      double operator ^(const Point &b)const{return x*b.y-y*b.x;}//叉积
      double operator *(const Point &b)const{return x*b.x + y*b.y;}//点积
      void transXY(double B){double tx = x,ty = y,x = tx*cos(B) - ty*sin(B),y = tx*sin(B) + ty*cos(B);} //绕原点旋转角度B（弧度值），后x,y的变化
 }list[MAXN];
 int S[MAXN],top;//相对于list[0]的极角排序
 int sgn(double x) {
  if(fabs(x) < eps)return ;
  if(x < )return -;
  else return ;
 }
 double dist(Point a,Point b){return sqrt((a-b)*(a-b));}
 bool _cmp(Point p1,Point p2){
     double tmp =(p1-list[])^(p2-list[]);
     if(sgn(tmp)>)return ;
     else if(sgn(tmp)==&&sgn(dist(p1,list[])-dist(p2,list[]))<= )return ;
     return ;
 }
 void Graham(int n){
     Point p0=list[],tp;
     int k=;
     for(int i=;i<n;i++)if((p0.y > list[i].y)||(p0.y ==list[i].y&&p0.x>list[i].x))p0 =list[i],k=i;
     tp=list[k],list[k]=list[],list[]=tp,sort(list+,list+n,_cmp);
     if(n==){top=,S[]=;return;}
     if(n==){top=,S[]=,S[]=;return;}
     S[]=,S[]=,top=;
     for(int i=;i<n;i++){
         while(top>&&sgn((list[S[top-]]-list[S[top-]])^(list[i]-list[S[top-]]))<=)top--;
         S[top++]=i;
     }
 }

 int main(){
     int t,n,l;
     scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&l);
         F(i,,n-)scanf("%lf%lf",&list[i].x,&list[i].y);
         Graham(n);
         double ans=;
         F(i,,top-)ans+=dist(list[S[i]],list[S[i+]]);
         ans+=dist(list[S[]],list[S[top-]])+PI**l;
         printf("%.0lf\n",ans);
         if(t!=)puts("");
     }
     return ;
 }