Poj_1269 Intersecting Lines -判两直线状态（水题、坑OJ）-白红宇

Poj_1269 Intersecting Lines -判两直线状态（水题、坑OJ）

阅读量：5165 次

发布时间：2019-06-13

本文共 1873 字，大约阅读时间需要 6 分钟。

题意：判直线相交、平行、重合

吐槽：坑OJ，G++WA，C++AC，模版题

/************************************************Author        :DarkTongCreated Time  :2016/8/3 23:02:25File Name     :Poj_1269.cpp*************************************************///#include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;#define eps 1e-9typedef long long LL;typedef double Db;struct Point{    Db x, y;    Point(Db x=0, Db y=0):x(x), y(y){}}C;typedef Point Vector;//向量或点的四则运算Vector operator + (Vector A, Vector B){ return Vector(A.x+B.x, A.y+B.y);}Vector operator - (Vector A, Vector B){ return Vector(A.x-B.x, A.y-B.y);}Vector operator * (Vector A, Db p){ return Vector(A.x*p, A.y*p);}Vector operator / (Vector A, Db p){ return Vector(A.x/p, A.y/p);}int dcmp(Db x){    if(fabs(x)
          
           0(-90, 90), ==0(90或-90), <0((-180,90)或(90，180])*/Db Cross(Vector A, Vector B){ return A.x*B.y - A.y*B.x;}    /*叉积(判<180的角): ==0(共线), >0(逆时针), <0(顺时针)  */Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){    Vector u = P-Q;    double t = Cross(w, u)/Cross(v, w); //刚好时两个向量同向，所以不用取绝对值    return P+v*t;}/**************************************************************/Point poi[4];int main(){    int T, cas=1;    scanf("%d", &T);    puts("INTERSECTING LINES OUTPUT");    while(T--)    {        for(int i=0;i<4;++i) scanf("%lf%lf", &poi[i].x, &poi[i].y);        Vector v1 = poi[1]-poi[0], v2 = poi[3]-poi[2];        if(dcmp(Cross(v1, v2))==0)        {            if(dcmp(Cross(poi[0]-poi[2], v1))==0) puts("LINE");            else puts("NONE");        }        else         {            Point t = GetLineIntersection(poi[0], v1, poi[2], v2);            printf("POINT %.2lf %.2lf\n", t.x, t.y);        }    }    puts("END OF OUTPUT");        return 0;}