hdu 4998 Rotate 点的旋转 银牌题

Posted 2020-08-12 smilesundream

Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1232    Accepted Submission(s): 545
Special Judge

Problem Description

Noting is more interesting than rotation!

Your little sister likes to rotate things. To put it easier to analyze, your sister makes n rotations. In the i-th time, she makes everything in the plane rotate counter-clockwisely around a point ai by a radian of pi.

Now she promises that the total effect of her rotations is a single rotation around a point A by radian P (this means the sum of pi is not a multiplier of 2π).

Of course, you should be able to figure out what is A and P :).

 

 

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains an integer n denoting the number of the rotations. Then n lines follows, each containing 3 real numbers x, y and p, which means rotating around point (x, y) counter-clockwisely by a radian of p.

We promise that the sum of all p‘s is differed at least 0.1 from the nearest multiplier of 2π.

T<=100. 1<=n<=10. 0<=x, y<=100. 0<=p<=2π.

 

 

Output

For each test case, print 3 real numbers x, y, p, indicating that the overall rotation is around (x, y) counter-clockwisely by a radian of p. Note that you should print p where 0<=p<2π.

Your answer will be considered correct if and only if for x, y and p, the absolute error is no larger than 1e-5.

 

 

Sample Input

1 3 0 0 1 1 1 1 2 2 1

 

 

Sample Output

1.8088715944 0.1911284056 3.0000000000

 

 

Source

2014 ACM/ICPC Asia Regional Anshan Online

题意：在平面上给你n个点，每个点都有一个对应的弧度，问整个平面依次绕每个点逆时针旋转对应的弧度

最后相当于对哪个点旋转了多少弧度？

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <map>
#include <bitset>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>

#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define CT continue
#define SC scanf

const double eps=1e-8;
const double pi=acos(-1);
int cas,n;

struct Point{
   double x,y,pi;
   void read(){
      SC("%lf%lf%lf",&x,&y,&pi);
   }
}p[15];

int dcmp(double a)
{
   if(fabs(a)<eps) return 0;
   else return a>0?1:-1;
}

double dot(Point a,Point b)
{
    return a.x*b.x+a.y*b.y;
}

double Length(Point a)
{
   return sqrt(a.x*a.x+a.y*a.y);
}

Point operator*(double k,Point a)
{
   return  (Point){k*a.x,k*a.y,0};
}

Point operator-(Point a,Point b)
{
   return (Point){a.x-b.x,a.y-b.y,0};
}

Point operator+(Point a,Point b)
{
    return (Point){a.x+b.x,a.y+b.y,0};
}

Point Rotate(Point a,double rad)
{
   return (Point){a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad),0};
}

Point Normal(Point a)
{
    double L=Length(a);
    return (Point){-a.y/L,a.x/L,0};
}

void R(Point &s,Point &t,Point &nor,Point &mid)
{
    s=p[0];
    for(int i=1;i<=n;i++){
        Point v=p[0]-p[i];
        v=Rotate(v,p[i].pi);
        p[0]=p[i]+v;
    }
    t=p[0];
    nor=Normal(t-s),mid=(Point){(s.x+t.x)/2,(s.y+t.y)/2,0};
}

double cross(Point a,Point b)
{
    return a.x*b.y-b.x*a.y;
}

Point Getlineintersection(Point p,Point v,Point q,Point w)
{
    Point u=p-q;
    double t=cross(w,u)/cross(v,w);
    return p+t*v;
}

int main()
{
    SC("%d",&cas);
    while(cas--)
    {
        SC("%d",&n);
        for(int i=1;i<=n;i++) p[i].read();
        Point s1,t1,s2,t2,nor1,nor2,mid1,mid2,o;
        p[0]={12,9,0};
        R(s1,t1,nor1,mid1);
        p[0]={3,17,0};
        R(s2,t2,nor2,mid2);

        o=Getlineintersection(mid1,nor1,mid2,nor2);
        Point os=s1-o,ot=t1-o;

        double ang,ang1=atan2(os.y,os.x),ang2=atan2(ot.y,ot.x);
        if(dcmp(ang1-ang2)>0) ang=ang1-ang2;
        else ang=ang2-ang1;

        if(dcmp(cross(t1-s1,o-s1))>0){
           if(dcmp(ang-pi)>0) ang=2*pi-ang;
        }
        else {
           if(dcmp(pi-ang)>0) ang=2*pi-ang;
        }

        if(dcmp(o.x)==0) o.x=0;
        if(dcmp(o.y)==0) o.y=0;
        printf("%.10f %.10f %.10f\n",o.x,o.y,ang);
    }
    return 0;
}

　　1.因为题目保证有解，所以在平面上任取两点，求出旋转后的弧度，那么最后选装的圆心必定在两条

从起点到终点的中垂线上

      2.找到圆心后，再依据圆心在向量st的左侧还是右侧确定旋转的弧度是>pi还是小于pi。

      3.最后因为double表示数据时，0可能是2*1e-30，按%.10f输出则是-0.00000000，所以需要

在最后判断一下。