POJ 1269 判斷直線與直線相交

題目意思:

給出8個數值 = 4個座標 = 2條直線

問兩條直線的關係: 相交(交點), 共線,平行;

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <queue>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8const int maxn = 1000+3; int N;struct Point{    double x;    double y;    Point(double a = 0.0, double b = 0.0) { x = a; y = b; }};struct Line{    Point u;    Point v;} L1,L2;int Sig(double x){    return (x > eps) - (x < -eps);}double Mult(const Point &p0, const Point &p1, const Point &p2){    return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y);}double Rake(Line L)  //求斜率{    if(!Sig(L.v.x-L.u.x))        return INF;    return (L.v.y - L.u.y)/(L.v.x - L.u.x);}int Intersect(Line L1, Line L2) //line instersect in a point{    int a = Sig(Mult(L2.u, L1.u, L2.v));    int b = Sig(Mult(L2.u, L1.v, L2.v));    if( a*b<0) //相交        return 1;    else if( a*b == 0 ) //共線        return 0;    else  //相離  (包括平行)        return -1;} int main(){#ifndef ONLINE_JUDGE    freopen("in","r",stdin);#endif    cin>>N;    cout<<"INTERSECTING LINES OUTPUT"<<endl;    double k1,k2,b1,b2,x,y;    while(N--)    {        cin>>L1.u.x>>L1.u.y>>L1.v.x>>L1.v.y>>L2.u.x>>L2.u.y>>L2.v.x>>L2.v.y;        int flag = Intersect(L1, L2);        if(flag == 1)        {            k1 = Rake(L1);            b1 = L1.u.y - k1*L1.u.x;            k2 = Rake(L2);            b2 = L2.u.y - k2*L2.u.x;            x = (b2-b1) / (k1-k2);            y = k1*x + b1;            printf("POINT %.2f %.2f\n",x,y);        }        else if(flag == 0)            cout<<"LINE"<<endl;        else            cout<<"NONE"<<endl;    }    cout<<"END OF OUTPUT"<<endl;    return 0;} 

起初我還判斷了,斜率等於無窮的情況,後來發現,除以無窮大 就是0了...所以不需要判斷.  判斷代碼:

//            if(!Sig(INF-k1) || !Sig(INF-k2)) //存在其中一條直線垂直x軸//            {//                if(!Sig(INF-k1))  //L1 垂直x軸//                {//                    x = L1.u.x;//                    y = k2*x + b2;//                }//                else             //L2 垂直x軸//                {//                    x = L2.u.x;//                    y = k1*x + b1;//                }//            }//            else//            {            x = (b2-b1) / (k1-k2);            y = k1*x + b1;//            }