poj2194Stacking Cylinders

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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.

  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-10;

 15 const double pi = acos(-1.0);

 16 const double inf = ~0u>>2;

 17 const double r = 1.0;

 18 const double rd = 3.40;

 19 struct point

 20 {

 21     double x,y,r;

 22     point(double x=0,double y=0):x(x),y(y){}

 23 }p[N],q[N];

 24 typedef point pointt;

 25 pointt operator -(point a,point b)

 26 {

 27     return point(a.x-b.x,a.y-b.y);

 28 }

 29 double dis(point a)

 30 {

 31     return sqrt(a.x*a.x+a.y*a.y);

 32 }

 33 double dot(point a,point b)

 34 {

 35     return a.x*b.x+a.y*b.y;

 36 }

 37 double angle(point a,point b)

 38 {

 39     return acos(dot(a,b)/dis(a)/dis(b));

 40 }

 41 double angle1(point v)

 42 {

 43     return atan2(v.y,v.x);

 44 }

 45 int dcmp(double x)

 46 {

 47     if(fabs(x)<eps) return 0;

 48     return x<0?-1:1;

 49 }

 50 point rotate(point a,double rad)//逆时针旋转

 51 {

 52     return point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));

 53 }

 54 bool cmp(point a,point b)

 55 {

 56     return a.x<b.x;

 57 }

 58 int main()

 59 {

 60     int n,i,j;

 61     while(scanf("%d",&n)&&n)

 62     {

 63         for(i = 1; i <= n;i++)

 64         {

 65             double x;

 66             scanf("%lf",&x);

 67             p[i] = point(x,1);

 68         }

 69         sort(p+1,p+n+1,cmp);

 70         int tn = n;

 71         double maxz = 0;

 72         point tp;

 73         while(1)

 74         {

 75             int g = 0;

 76             for(i = 1; i < tn; i++)

 77             {

 78                 if(maxz<p[i].y)

 79                 {

 80                     maxz = p[i].y;

 81                     tp = p[i];

 82                 }

 83                // if(dcmp(dis(p[i]-p[i+1])-2*r)<0) continue;

 84                // if(dcmp(dis(p[i]-p[i+1])-rd)>0) continue;

 85 

 86                 double b = atan2(fabs(p[i].y-p[i+1].y),fabs(p[i].x-p[i+1].x));

 87                 double d = dis(p[i]-p[i+1])/2;

 88                 double dd = sqrt(4-(d*d));

 89                 double a = atan2(dd,d);

 90                 if(dcmp(a+b-pi/2.0)>0) continue;

 91                 if(p[i].y<p[i+1].y)

 92                 {

 93                     point pp = point(2.0,0);

 94                     q[++g] = rotate(pp,a+b);

 95                     q[g] = point(q[g].x+p[i].x,q[g].y+p[i].y);

 96                 }

 97                 else

 98                 {

 99                     point pp = point(-2,0);

100                     q[++g] = rotate(pp,-a-b);

101                     q[g] = point(q[g].x+p[i+1].x,q[g].y+p[i+1].y);

102                 }

103             }

104             if(maxz<p[tn].y)

105             {

106                 maxz  = p[tn].y;

107                 tp = p[tn];

108             }

109             if(!g) break;

110             tn = g;

111             for(i = 1; i <= g ; i++)

112             p[i] = q[i];

113         }

114         printf("%.4f %.4f\n",tp.x,tp.y);

115     }

116     return 0;

117 }

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