• c++实现 给定直角停车位两个点,求取剩余两点坐标。


    //2018-09-08-fourmi
    
    
    /*************************include head files************************************************/
    #include<iostream>
    #include<math.h>
    #include<vector>
    #include<cmath>
    #include<iomanip>
    #include<cstdlib>
    #include<cstring>
    #include<fstream>
    /*******************************************************************************************/
    
    
    
    
    /**********************Define variables*****************************************************/
    #define pi 3.1415926
    #define POINT   pair<double,double>  
    #define VECT vector<vector<double> >
    const int MAX_ITER=100000;
    const double eps=0.0000001;
    /*******************************************************************************************/
    
    
    
    
    using namespace std;
    
    
    
    
    
    /*****************************Vector_Transformation****************************************/
    POINT Vector_Transformation(POINT &pt,double num1,double num2,double num3,double num4)
    {
       POINT result;
       result.first=pt.first*num1+pt.second*num3;
       result.second=pt.first*num2+pt.second*num4;
       
       return result;
       
    }
    /*******************************************************************************************/
    
    
    
    
    /***************************FUNCTIONS_ABOUT_SVD*********************************************/
    double get_norm(double *x, int n){
        double r=0;
    
        for(int i=0;i<n;i++)
            r+=x[i]*x[i];
        return sqrt(r);
    }
    
    double normalize(double *x, int n){
        double r=get_norm(x,n);
    
        if(r<eps)
            return 0;
        for(int i=0;i<n;i++)
            x[i]/=r;
        return r;
    }
    
    inline double product(double*a, double *b,int n){
        double r=0;
    
        for(int i=0;i<n;i++)
            r+=a[i]*b[i];
        return r;
    }
    
    void orth(double *a, double *b, int n){//|a|=1
        double r=product(a,b,n);
     
        for(int i=0;i<n;i++)
            b[i]-=r*a[i];
        
    }
    
    bool svd(VECT A, int K, VECT &U, vector<double> &S, VECT &V){
        int M=A.size();
        int N=A[0].size();
        double *left_vector=new double[M];
        double *next_left_vector=new double[M];
        double *right_vector=new double[N];
        double *next_right_vector=new double[N];
        double diff=1;
        double r=-1;
        int col=0;
    
        U.clear();
        V.clear();
        S.clear();
        S.resize(K,0);
        U.resize(K);
        for(int i=0;i<K;i++)
            U[i].resize(M,0);
        V.resize(K);
        for(int i=0;i<K;i++)
            V[i].resize(N,0);    
    
        for(int col=0;col<K;col++){
    
            while(1){
                for(int i=0;i<M;i++)
                    left_vector[i]= (float)rand() / RAND_MAX;
                if(normalize(left_vector, M)>eps)
                    break;
                    }
     
            for(int iter=0;diff>=eps && iter<MAX_ITER;iter++){
                memset(next_left_vector,0,sizeof(double)*M);
                memset(next_right_vector,0,sizeof(double)*N);
                for(int i=0;i<M;i++)
                    for(int j=0;j<N;j++)
                        next_right_vector[j]+=left_vector[i]*A[i][j];
     
                r=normalize(next_right_vector,N);
                if(r<eps) break;
                for(int i=0;i<col;i++)
                    orth(&V[i][0],next_right_vector,N);
                normalize(next_right_vector,N);
     
                for(int i=0;i<M;i++)
                    for(int j=0;j<N;j++)
                        next_left_vector[i]+=next_right_vector[j]*A[i][j];
                r=normalize(next_left_vector,M);
                if(r<eps) break;
                for(int i=0;i<col;i++)
                    orth(&U[i][0],next_left_vector,M);
                normalize(next_left_vector,M);
                diff=0;
                
                for(int i=0;i<M;i++){
                    double d=next_left_vector[i]-left_vector[i];
                    diff+=d*d;
                }
     
                memcpy(left_vector,next_left_vector,sizeof(double)*M);
                memcpy(right_vector,next_right_vector,sizeof(double)*N);
            }
            
            if(r>=eps){
                S[col]=r;
                memcpy((char *)&U[col][0],left_vector,sizeof(double)*M);
                memcpy((char *)&V[col][0],right_vector,sizeof(double)*N);
            }else{
                cout<<r<<endl;
                break;
            }
        }
       
        delete [] next_left_vector;
        delete [] next_right_vector;
        delete [] left_vector;
        delete [] right_vector;
     
        return true;
    }
    /*******************************************************************************************/
    
    
    
    
    /**********************GET_THE_BIGGEST_SINGULAR_VALUE***************************************/
    vector<double> GET_THE_BIGGEST_SINGULAR_VALUE(POINT vec,int m,int n,int k)
    {
        //分解一个1*2的矩阵A,求其前1个奇异值和奇异向量
        
        VECT A;
        A.resize(m);
        
        for(int i=0;i<m;i++)
          {
            A[i].resize(n);
          }
        A[0][0]=vec.first;
        A[0][1]=vec.second; 
    
        VECT U;
        vector<double> S;
        VECT V;
        svd(A,k,U,S,V);
       return S;
    }
    /*******************************************************************************************/
    
    
    
    
    /************************CAL_THIRD_AND_FORTH_POINTS*****************************************/
    POINT * cal_reset_TWO_points(POINT &pt1,POINT &pt2)
    {
       static POINT arr[2];
       POINT vecFromSecToFirst,vec,pt3,pt4;
       int x1,y1,x2,y2,exemplarStart,exemplarEnd;
       double distance,angle,s,sideLength;
       double slantAngleInRadians = (double(90)/180)*pi;
       double LongSideMin=47.9042;
       double verticalPSSideLength=195;
       double parallelPSSideLength=83;
       int matrix_rows=1;
       int matrix_cols=2;
       int top_k_max=1;
      
       x1=pt1.first;
       y1=pt1.second;
       x2=pt2.first;
       y2=pt2.second;
    
       distance = sqrt(pow((x1-x2),2)+pow((y1-y2),2));
    
       if (distance < LongSideMin) //说明是短库位
          sideLength = verticalPSSideLength;
       else //说明是平行长库位
          sideLength = parallelPSSideLength;
       
       vecFromSecToFirst.first=pt1.first-pt2.first;
       vecFromSecToFirst.second=pt1.second-pt2.second;
       vec=Vector_Transformation(vecFromSecToFirst,cos(slantAngleInRadians),-sin(slantAngleInRadians), sin(slantAngleInRadians), cos(slantAngleInRadians));
    
    
       if ((pt2.first-pt1.first)==0)
        {
           angle=pi/2;
        }
        else
        {
           angle=abs(atan2(vecFromSecToFirst.second,vecFromSecToFirst.first));      
        }
    
        s=GET_THE_BIGGEST_SINGULAR_VALUE(vec,matrix_rows,matrix_cols,top_k_max)[0];   
        vec.first=vec.first/s;
        vec.second=vec.second/s;
    
        pt3.first=pt2.first+vec.first*sideLength;
        pt3.second=pt2.second+vec.second*sideLength;
        pt4.first=pt1.first+vec.first*sideLength;
        pt4.second=pt1.second+vec.second*sideLength;
        arr[0]=pt3;
        arr[1]=pt4;
    
    
       return arr;
    }
    /*******************************************************************************************/
    
    
    
    
    /*****************************************MAIN()********************************************/
    int main()
    {
    
      POINT * arr;
      POINT pt1(171,145);
      POINT pt2(171,213);
      arr=cal_reset_TWO_points(pt1,pt2);
      std::cout<<arr[0].first<<"  "<<arr[0].second<<std::endl;
      std::cout<<arr[1].first<<"  "<<arr[1].second<<std::endl;
    
      return 0;
      
    }
    /*******************************************************************************************/
    

      

  • 相关阅读:
    c++父类指针子类指针转化分析
    setbuf手册
    c++细节
    cf727e
    总结
    UVa 10192 Vacation (最长公共子序列)
    HUNNU 11313 最长公共子序列(LCS)
    HDU 2069 Coin Change (经典DP)
    UVa 674 Coin Change (经典DP)
    UVa 10131 Is Bigger Smarter? (LDS+数据结构排序)
  • 原文地址:https://www.cnblogs.com/fourmi/p/9619849.html
Copyright © 2020-2023  润新知