hdu2297Run（凸包的巧妙应用）

链接

很巧妙的一道题，参考

把距离和速度分别作为x和y坐标，以斜率代表追赶速率，简直炫酷~

具体看上面的博客，画的很清楚，就不再抄写一遍了。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 50010
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct point
 {
     double x,y;
     point(double x=,double y = ):x(x),y(y){}
 }p[N],ch[N];
 typedef point pointt;
 point operator -(point a,point b)
 {
     return point(a.x-b.x,a.y-b.y);
 }
 double cross(point a,point b)
 {
     return a.x*b.y-a.y*b.x;
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
      else return x<?-:;
 }
 double mul(point p0,point p1,point p2)
 {
     return cross(p1-p0,p2-p0);
 }
 double dis(point a)
 {
     return sqrt(a.x*1.0*a.x+a.y*1.0*a.y);
 }
 bool cmp(point a,point b)
 {
     if(dcmp(mul(p[],a,b))==)
         return dis(a-p[])<dis(b-p[]);
     else
         return dcmp(mul(p[],a,b))>;
 }
 int Graham(int n)
 {
     if(n<) return n;
     int i,k = ,top;
     point tmp;
     for(i =  ; i < n; i++)
     {
         if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x))
             k = i;
     }
     if(k!=)
     {
         tmp = p[];
         p[] = p[k];
         p[k] = tmp;
     }
     sort(p+,p+n,cmp);
     ch[] = p[];
     ch[] = p[];
     top = ;
     for(i = ; i < n ; i++)
     {
         while(top>&&dcmp(mul(ch[top-],ch[top],p[i]))<=)
             top--;
         top++;
         ch[top] = p[i];
     }
     return top;
 }
 int main()
 {
     int t,i,n;
     cin>>t;
     while(t--)
     {
         scanf("%d",&n);
         for(i =  ; i < n; i++)
         scanf("%lf%lf",&p[i].y,&p[i].x);
         int m = Graham(n);
         int maxx = -INF,tx,maxy = -INF,ty;
         if(n<)
         {
             cout<<"1\n";
             continue;
         }
         for(i =  ; i <= m ; i++)
         {
             //cout<<ch[i].x<<" "<<ch[i].y<<endl;
             if(ch[i].x>maxx)
             {
                 tx = i;
                 maxx = ch[i].x;
             }
             if(ch[i].y>maxy)
             {
                 ty = i;
                 maxy = ch[i].y;
             }
         }
         cout<<ty-tx+<<endl;
     }
     return ;
 }