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-10;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
const double r = 1.0;
const double rd = 3.40;
struct point
{
    double x,y,r;
    point(double x=0,double y=0):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 0;
    return x<0?-1:1;
}
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 = 1; i <= n;i++)
        {
            double x;
            scanf("%lf",&x);
            p[i] = point(x,1);
        }
        sort(p+1,p+n+1,cmp);
        int tn = n;
        double maxz = 0;
        point tp;
        while(1)
        {
            int g = 0;
            for(i = 1; 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+1].y),fabs(p[i].x-p[i+1].x));
                double d = dis(p[i]-p[i+1])/2;
                double dd = sqrt(4-(d*d));
                double a = atan2(dd,d);
                if(dcmp(a+b-pi/2.0)>0) continue;
                if(p[i].y<p[i+1].y)
                {
                    point pp = point(2.0,0);
                    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(-2,0);
                    q[++g] = rotate(pp,-a-b);
                    q[g] = point(q[g].x+p[i+1].x,q[g].y+p[i+1].y);
                }
            }
            if(maxz<p[tn].y)
            {
                maxz  = p[tn].y;
                tp = p[tn];
            }
            if(!g) break;
            tn = g;
            for(i = 1; i <= g ; i++)
            p[i] = q[i];
        }
        printf("%.4f %.4f
",tp.x,tp.y);
    }
    return 0;
}