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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