• uva 1453


    旋转卡壳算法;

    直接在这个上面粘的模板

    主要用途:用于求凸包的直径、宽度,两个不相交凸包间的最大距离和最小距离···

    这题就是求凸包的直径

      1 #include <cstdio>
      2 #include <cmath>
      3 #include <cstring>
      4 #include <algorithm>
      5 #include <vector>
      6 #define eps 1e-9
      7 using namespace std;
      8 const double pi = acos(-1);
      9 
     10 int dcmp(double x)
     11 {
     12     return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);
     13 }
     14 
     15 struct Point
     16 {
     17     double x;
     18     double y;
     19 
     20     Point(double x = 0, double y = 0):x(x), y(y) {}
     21 
     22     bool operator < (const Point& e) const
     23     {
     24         return dcmp(x - e.x) < 0 || (dcmp(x - e.x) == 0 && dcmp(y - e.y) < 0);
     25     }
     26 
     27     bool operator == (const Point& e) const
     28     {
     29         return dcmp(x - e.x) == 0 && dcmp(y - e.y) == 0;
     30     }
     31 };
     32 
     33 typedef Point Vector;
     34 
     35 Vector operator + (Point A, Point B)
     36 {
     37     return Vector(A.x + B.x, A.y + B.y);
     38 }
     39 
     40 Vector operator - (Point A, Point B)
     41 {
     42     return Vector(A.x - B.x, A.y - B.y);
     43 }
     44 
     45 Vector operator * (Point A, double p)
     46 {
     47     return Vector(A.x * p, A.y * p);
     48 }
     49 
     50 Vector operator / (Point A, double p)
     51 {
     52     return Vector(A.x / p, A.y / p);
     53 }
     54 double dot(Point a,Point b)
     55 {
     56     return a.x*b.x+a.y*b.y;
     57 }
     58 double cross(Point a,Point b)
     59 {
     60     return a.x*b.y-a.y*b.x;
     61 }
     62 Point rotate(Point a,double ang)
     63 {
     64     return Point(a.x*cos(ang)-a.y*sin(ang),a.x*sin(ang)+a.y*cos(ang));
     65 }
     66 int convexhull(Point *p,int n,Point *ch)
     67 {
     68     sort(p,p+n);
     69     int m=0;
     70     for(int i=0; i<n; i++)
     71     {
     72         while(m>1&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
     73         ch[m++]=p[i];
     74     }
     75     int k=m;
     76     for(int i=n-2; i>=0; i--)
     77     {
     78         while(m>k&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
     79         ch[m++]=p[i];
     80     }
     81     if(n>1)m--;
     82     return m;
     83 }
     84 bool onsegment(Point p,Point a,Point b)
     85 {
     86     return dcmp(cross(a-p,b-p))==0&&dcmp(dot(a-p,b-p))<0;
     87 }
     88 
     89 
     90 bool SegmentProperIntersection( Point a1, Point a2, Point b1, Point b2 )  //线段相交,交点不在端点
     91 {
     92     double c1 = cross( a2 - a1, b1 - a1 ), c2 = cross( a2 - a1, b2 - a1 ),
     93                 c3 = cross( b2 - b1, a1 - b1 ), c4 = cross( b2 - b1, a2 - b1 );
     94     return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
     95 }
     96 
     97 int ispointinpolygon(Point p,int n,Point *poly)
     98 {
     99     int wn=0;
    100     for(int i=0; i<n; i++)
    101     {
    102         if(onsegment(p,poly[i],poly[(i+1)%n]))return -1;
    103         int k=dcmp(cross(poly[(i+1)%n]-poly[i],p-poly[i]));
    104         int d1=dcmp(poly[i].y-p.y);
    105         int d2=dcmp(poly[(i+1)%n].y-p.y);
    106         if(k>0&&d1<=0&&d2>0)wn++;
    107         if(k<0&&d2<=0&&d1>0)wn--;
    108     }
    109     if(wn!=0)return 1;
    110     return 0;
    111 }
    112 
    113 bool Check(int n,Point *ch,int m,Point *th)
    114 {
    115     for(int i=0; i<n; i++)
    116     {
    117         if(ispointinpolygon(ch[i],m,th)!=0)return 1;
    118     }
    119     for(int i=0; i<m; i++)
    120         if(ispointinpolygon(th[i],n,ch)!=0)return 1;
    121     ch[n]=ch[0];
    122     th[m]=th[0];
    123     for(int i=0; i<n; i++)
    124         for(int j=0; j<m; j++)
    125             if(SegmentProperIntersection(ch[i],ch[i+1],th[j],th[j+1]))return 1;
    126     return 0;
    127 }
    128 double rotating_calipers(Point *ch,int n)
    129 {
    130     int q=1;
    131     double ans=0;
    132     ch[n]=ch[0];
    133     for ( int i = 0; i < n; ++i )
    134      {
    135          while ( cross( ch[i + 1] - ch[i], ch[q + 1] - ch[i] ) > cross( ch[i + 1] - ch[i], ch[q] - ch[i] ) )
    136             q = ( q + 1 ) % n;
    137          ans = max( ans, max( dot( ch[i]- ch[q],ch[i]-ch[q] ),dot( ch[i + 1]-ch[q + 1],ch[i + 1]-ch[q + 1] ) ));
    138     }
    139     return ans;
    140 }
    141 Point p[400009],ch[400009];
    142 int main()
    143 {
    144     int n,m;
    145     double x,y,w;
    146     int t;
    147     scanf("%d",&t);
    148     while(t--)
    149     {
    150         scanf("%d",&n);
    151         int cnt=0;
    152         for(int i=0; i<n; i++)
    153         {
    154             scanf("%lf%lf%lf",&x,&y,&w);
    155             p[cnt].x=x,p[cnt++].y=y;
    156             p[cnt].x=x+w,p[cnt++].y=y;
    157             p[cnt].x=x,p[cnt++].y=y+w;
    158             p[cnt].x=x+w,p[cnt++].y=y+w;
    159         }
    160         int n1=convexhull(p,cnt,ch);
    161         printf("%.0lf
    ",rotating_calipers(ch,n1));
    162     }
    163     return 0;
    164 }
    View Code
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  • 原文地址:https://www.cnblogs.com/yours1103/p/3407348.html
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