poj2194Stacking Cylinders

链接

可以根据反余弦和反正切算出角a和b的值， 然后向量旋转就可以了，图中的状态旋转rotate（(2,0),a+b)  反状态把角度反过来，点取(-2,0)即可。

不知道是不是理解错了，题意写着两圆距离》2，《3.4，在求得时候就加了特判，一直WA。。。去了特判就过了。

为了提高精度，可以全化为atan2.

 #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 double r = 1.0;
 const double rd = 3.40;
 struct point
 {
     double x,y,r;
     point(double x=,double y=):x(x),y(y){}
 }p[N],q[N];
 typedef point pointt;
 pointt operator -(point a,point b)
 {
     return point(a.x-b.x,a.y-b.y);
 }
 double dis(point a)
 {
     return sqrt(a.x*a.x+a.y*a.y);
 }
 double dot(point a,point b)
 {
     return a.x*b.x+a.y*b.y;
 }
 double angle(point a,point b)
 {
     return acos(dot(a,b)/dis(a)/dis(b));
 }
 double angle1(point v)
 {
     return atan2(v.y,v.x);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }
 point rotate(point a,double rad)//逆时针旋转
 {
     return point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
 }
 bool cmp(point a,point b)
 {
     return a.x<b.x;
 }
 int main()
 {
     int n,i,j;
     while(scanf("%d",&n)&&n)
     {
         for(i = ; i <= n;i++)
         {
             double x;
             scanf("%lf",&x);
             p[i] = point(x,);
         }
         sort(p+,p+n+,cmp);
         int tn = n;
         double maxz = ;
         point tp;
         while()
         {
             int g = ;
             for(i = ; i < tn; i++)
             {
                 if(maxz<p[i].y)
                 {
                     maxz = p[i].y;
                     tp = p[i];
                 }
                // if(dcmp(dis(p[i]-p[i+1])-2*r)<0) continue;
                // if(dcmp(dis(p[i]-p[i+1])-rd)>0) continue;

                 double b = atan2(fabs(p[i].y-p[i+].y),fabs(p[i].x-p[i+].x));
                 double d = dis(p[i]-p[i+])/;
                 double dd = sqrt(-(d*d));
                 double a = atan2(dd,d);
                 if(dcmp(a+b-pi/2.0)>) continue;
                 if(p[i].y<p[i+].y)
                 {
                     point pp = point(2.0,);
                     q[++g] = rotate(pp,a+b);
                     q[g] = point(q[g].x+p[i].x,q[g].y+p[i].y);
                 }
                 else
                 {
                     point pp = point(-,);
                     q[++g] = rotate(pp,-a-b);
                     q[g] = point(q[g].x+p[i+].x,q[g].y+p[i+].y);
                 }
             }
             if(maxz<p[tn].y)
             {
                 maxz  = p[tn].y;
                 tp = p[tn];
             }
             if(!g) break;
             tn = g;
             for(i = ; i <= g ; i++)
             p[i] = q[i];
         }
         printf("%.4f %.4f\n",tp.x,tp.y);
     }
     return ;
 }