1393: Robert Hood 旋转卡壳 凸包

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1393

http://poj.org/problem?id=2187 Beauty Contest

1393: Robert Hood

Description

Input

Output

Sample Input

5
-4 1
-100 0
0 4
2 -3
2 300

Sample Output

316.86590223

HINT

Source

分析：

给你 N 个点, 求所有点中最远两点距离。即是凸包直径。

凸包+旋转卡壳

AC代码：

 #include<stdio.h>
 #include<math.h>
 #include<string.h>
 #include<algorithm>
 using namespace std;

 const int maxn = +;
 int n,m;

 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);
     }
 }p[maxn], ch[maxn];

 bool cmp(Point p1, Point p2)
 {
     if(p1.x == p2.x) return p1.y < p2.y;
     return p1.x < p2.x;
 }

 int squarDist(Point A, Point B) /**距离的平方*/
 {
     return (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);
 }

 double Cross(Point A, Point B) /**叉积*/
 {
     return A.x*B.y-A.y*B.x;
 }

 void ConvexHull() /** 基于水平的Andrew算法求凸包 */
 {
     sort(p,p+n,cmp); /**先按照 x 从小到大排序, 再按照 y 从小到大排序*/
     m = ;

     for(int i = ; i < n; i++) /** 从前往后找 */
     {
         while(m >  && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
         ch[m++] = p[i];
     }
     int k = m;
     for(int i = n-; i >= ; i--) /**从后往前找, 形成完整的封闭背包*/
     {
         while(m > k && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
         ch[m++] = p[i];
     }
     if(n > ) m--;
 }

 int rotating_calipers() /**旋转卡壳模板*/
 {
     int q = ;
     int ans = ;
     ch[m] = ch[]; /**凸包边界处理*/
     for(int i = ; i < m; i++) /**依次用叉积找出凸包每一条边对应的最高点*/
     {
         while(Cross(ch[i+]-ch[i], ch[q+]-ch[i]) > Cross(ch[i+]-ch[i], ch[q]-ch[i]))
             q = (q+)%m;
         ans = max(ans, max(squarDist(ch[i], ch[q]), squarDist(ch[i+], ch[q+])));
     }
     return ans;
 }

 int main()
 {
     while(scanf("%d", &n) != EOF)
     {
         if(n == ) break;
         for(int i = ; i < n; i++)
             scanf("%lf%lf", &p[i].x, &p[i].y);

         ConvexHull();

         printf("%.8lf\n", sqrt(rotating_calipers()));
     }
     return ;
 }

同学用结构体 + 遍历凸包也可以解决。

AC代码：

 #include<iostream>
 #include<algorithm>
 #include<math.h>
 using namespace std;
 struct point
 {
     int x;
     int y;
 }p[],res[];
 int cmp(point p1,point p2)
 {
     return p1.y<p2.y||(p1.x==p2.x&&p1.x<p2.x);
 }
 bool ral(point p1,point p2,point p3)
 {
     return (p2.x-p1.x)*(p3.y-p1.y)>(p3.x-p1.x)*(p2.y-p1.y);
 }
 int main()
 {
     int n,i,j;
     while((scanf("%d",&n))!=EOF)
     {
         for(i=;i<n;i++)
             scanf("%d %d",&p[i].x,&p[i].y);
         sort(p,p+n,cmp);
         res[]=p[];
         res[]=p[];
         int top=;
         for(i=;i<n;i++)
         {
             while(top&&!ral(res[top],res[top-],p[i]))
                 top--;
             res[++top]=p[i];
         }
         int len=top;
         res[++top]=p[n-];
         for(i=n-;i>=;i--)
         {
             while(top!=len&&!ral(res[top],res[top-],p[i]))
                 top--;
             res[++top]=p[i];
         }
         double maxx=;
         for(i=;i<top;i++)
         {
             for(j=i+;j<top;j++)
             {
                 double s=(res[i].x-res[j].x)*(res[i].x-res[j].x)+(res[i].y-res[j].y)*(res[i].y-res[j].y);
                 if(s>maxx)
                     maxx=s;

             }
         }
         printf("%.8lf\n",sqrt(maxx));
     }
     return ;
 }