hdu3847Trash Removal（凸包）

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这题居然是WF的题， 应属于签到题。。

求一个多边形是否能被一个宽为d的矩形框住。

可以求一下凸包，然后枚举每条凸包的边，找出距离最远的点。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 100000
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 const int MAXN=;
 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);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }
 double dis(point a,point b)
 {
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }
 double cross(point a,point b)
 {
     return a.x*b.y-a.y*b.x;
 }
 double mul(point p0,point p1,point p2)
 {
     return cross(p1-p0,p2-p0);
 }
 double dis(point a)
 {
     return sqrt(a.x*a.x+a.y*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)
 {
     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;
 }
 double distancetoline(point p,point a,point b)
 {
     point v1 = b-a,v2 = p-a;
     return fabs(cross(v1,v2))/dis(v1);
 }
 int main()
 {
     int i,j;
     int n,kk=;
     while(scanf("%d",&n)&&n)
     {
         for(i =  ; i < n; i++)
         scanf("%lf%lf",&p[i].x,&p[i].y);
         int m = Graham(n);
         ch[m+] = ch[];
         double ans = INF;
         for(i =  ; i <= m; i++)
         {
             double ts = ;
             for(j =  ; j <= m ; j++)
             ts = max(ts,distancetoline(ch[j],ch[i],ch[i+]));
             ans = min(ans,ts);
         }
         ans = ceil(ans*)/;
         printf("Case %d: %.2lf\n",++kk,ans);
     }
     return ;
 }