poj 1269 Intersecting Lines

題意：給你兩條直線判斷是平行、重合還是相交，若相交還需輸出交點。對於每條直線只給出直線上兩點。

分析：用向量的叉積判斷平行和重合的情況，交點直接解方程求，斜率不存在特殊考慮。需要注意的是，有可能給的兩條直線中有相同的點，不注意的話，判斷共線的時候可能會出錯。

以下附上代碼：

#include <algorithm>
#include <iostream>
#include <sstream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cctype>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <map>
#include <set>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;


struct Point{
  double x,y;
  
  Point(){}
  Point(double x_, double y_) :
    x(x_), y(y_) {}
    
  //Point
  Point operator - (const Point &b) const{
    Point t(x-b.x,y-b.y);
    return t;
  }
  
  //Vector
  double operator ^ (const Point &b) const{
    return x*b.y - y*b.x;
  }
};

struct Line{
  Point p;
  double k;
  bool exist; //斜率存不存在 
  
  Line(){}
  Line(Point p_, double k_) :
    p(p_), k(k_) {}
  Line(Point a, Point b){
    p = a;
    if(a.x != b.x){
      k = (b.y-a.y)/(b.x-a.x);
      exist = 1;
    }
    else exist = 0;
      
  }
};


int n;
Point p1,p2;
Point p3,p4;


bool onLine(Point a, Point b, Point c)
{
  Point v1 = b-a;
  Point v2 = c-b;
  return (v1^v2) == 0;
}

bool parallel(Point a, Point b, Point c, Point d)
{
  Point v1 = b-a;
  Point v2 = d-c;
  return (v1^v2) == 0;
}

Point intersect(Line a, Line b)
{
  Point t;
  if(a.exist && b.exist){
    t.x = (a.k*a.p.x - b.k*b.p.x + b.p.y - a.p.y) / (a.k - b.k);
    t.y = a.k*(t.x-a.p.x) + a.p.y;
  }
  else{
    if(!a.exist){
      t.x = a.p.x;
      t.y = b.k*(t.x-b.p.x) + b.p.y;
    }
    else if(!b.exist){
      t.x = b.p.x;
      t.y = a.k*(t.x-a.p.x) + a.p.y; 
    }
  }
  return t;
}

int main()
{
  while(scanf("%d",&n) != EOF){
    puts("INTERSECTING LINES OUTPUT");
    for(int i = 0; i < n; i++){
      scanf("%lf%lf",&p1.x,&p1.y);
      scanf("%lf%lf",&p2.x,&p2.y);
      scanf("%lf%lf",&p3.x,&p3.y);
      scanf("%lf%lf",&p4.x,&p4.y);
      
      if(onLine(p1,p2,p3) && onLine(p1,p2,p4)){//!!!可能有點重合 
        puts("LINE");
      }
      else{
        Line l1(p1,p2);
        Line l2(p3,p4);
        if(parallel(p1,p2,p3,p4)){
          puts("NONE");
        }
        else{
          Point p = intersect(l1,l2);
          printf("POINT %.2lf %.2lf\n",p.x,p.y);
        }
      }
    }//end of for
    puts("END OF OUTPUT");
  }
	return 0;
}