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。

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<stdlib.h>
  6 #include<vector>
  7 #include<cmath>
  8 #include<queue>
  9 #include<set>
 10 using namespace std;
 11 #define N 100000
 12 #define LL long long
 13 #define INF 0xfffffff
 14 const double eps = 1e-8;
 15 const double pi = acos(-1.0);
 16 const double inf = ~0u>>2;
 17 const int MAXN=1550;
 18 struct point
 19 {
 20     double x,y;
 21     point(double x=0,double y=0):x(x),y(y){}
 22 }p[N],ch[N];
 23 typedef point pointt;
 24 point operator -(point a,point b)
 25 {
 26     return point(a.x-b.x,a.y-b.y);
 27 }
 28 int dcmp(double x)
 29 {
 30     if(fabs(x)<eps) return 0;
 31     return x<0?-1:1;
 32 }
 33 double dis(point a,point b)
 34 {
 35     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 36 }
 37 double cross(point a,point b)
 38 {
 39     return a.x*b.y-a.y*b.x;
 40 }
 41 double mul(point p0,point p1,point p2)
 42 {
 43     return cross(p1-p0,p2-p0);
 44 }
 45 double dis(point a)
 46 {
 47     return sqrt(a.x*a.x+a.y*a.y);
 48 }
 49 bool cmp(point a,point b)
 50 {
 51     if(dcmp(mul(p[0],a,b))==0)
 52         return dis(a-p[0])<dis(b-p[0]);
 53     else
 54         return dcmp(mul(p[0],a,b))>0;
 55 }
 56 int Graham(int n)
 57 {
 58     int i,k = 0,top;
 59     point tmp;
 60     for(i = 0 ; i < n; i++)
 61     {
 62         if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x))
 63             k = i;
 64     }
 65     if(k!=0)
 66     {
 67         tmp = p[0];
 68         p[0] = p[k];
 69         p[k] = tmp;
 70     }
 71     sort(p+1,p+n,cmp);
 72     ch[0] = p[0];
 73     ch[1] = p[1];
 74     top = 1;
 75     for(i = 2; i < n ; i++)
 76     {
 77         while(top>0&&dcmp(mul(ch[top-1],ch[top],p[i]))<0)
 78             top--;
 79         top++;
 80         ch[top] = p[i];
 81     }
 82     return top;
 83 }
 84 double distancetoline(point p,point a,point b)
 85 {
 86     point v1 = b-a,v2 = p-a;
 87     return fabs(cross(v1,v2))/dis(v1);
 88 }
 89 int main()
 90 {
 91     int i,j;
 92     int n,kk=0;
 93     while(scanf("%d",&n)&&n)
 94     {
 95         for(i = 0 ; i < n; i++)
 96         scanf("%lf%lf",&p[i].x,&p[i].y);
 97         int m = Graham(n);
 98         ch[m+1] = ch[0];
 99         double ans = INF;
100         for(i = 0 ; i <= m; i++)
101         {
102             double ts = 0;
103             for(j = 0 ; j <= m ; j++)
104             ts = max(ts,distancetoline(ch[j],ch[i],ch[i+1]));
105             ans = min(ans,ts);
106         }
107         ans = ceil(ans*100)/100;
108         printf("Case %d: %.2lf\n",++kk,ans);
109     }
110     return 0;
111 }
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posted @ 2014-08-31 15:28  _雨  阅读(292)  评论(0)  编辑  收藏  举报