POJ 1269 Intersecting Lines

传送门：点击打开链接

Intersecting Lines

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16736 Accepted: 7213

Description

We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect. 
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000. 

Input

The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).

Output

There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".

Sample Input

50 0 4 4 0 4 4 05 0 7 6 1 0 2 35 0 7 6 3 -6 4 -32 0 2 27 1 5 18 50 3 4 0 1 2 2 5

Sample Output

INTERSECTING LINES OUTPUTPOINT 2.00 2.00NONELINEPOINT 2.00 5.00POINT 1.07 2.20END OF OUTPUT

Source

Mid-Atlantic 1996

题意就是给你四个点，每两点确定一条直线，问两直线是否有交点，有交点输出，没有输出共线或平行。

关键在于判断斜率，因为有斜率不存在的情况，如果是单独求k的话，斜率不存在还要单独拿出来考虑，第一次就没判断k，各种WA。然后可以利用向量的叉乘，若叉乘为0代表两直线平行，再去判断是否共线，这样成功避开了k的取值。最后得到正确结果还是花样WA，竟然是因为G++的问题，看了一眼讨论版才知道，可怕啊。

代码实现:

#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<queue>#include<cstdio>#define ll int#define mset(a,x) memset(a,x,sizeof(a))using namespace std;const double PI=acos(-1);const int inf=0x3f3f3f3f;const double eps=1e-8;const int maxn=5e5+5;const int mod=1e9+7;int dir[4][2]={0,1,1,0,0,-1,-1,0};struct point{double x,y;point(){};point(double newx,double newy){x=newx;y=newy;}point operator -(const point &b){return point(x-b.x,y-b.y);}};struct line{point a,b;};int across(line p,line q){point tempa=(p.b-p.a);point tempb=(q.b-q.a);if(tempa.x*tempb.y-tempb.x*tempa.y==0)return 0;elsereturn 1;}int main(){int n,i,j,k;line a,b;cin>>n;cout<<"INTERSECTING LINES OUTPUT"<<endl;while(n--){cin>>a.a.x>>a.a.y>>a.b.x>>a.b.y>>b.a.x>>b.a.y>>b.b.x>>b.b.y;if(!across(a,b))        //平行 {double b1=a.a.y-a.a.x*((a.b.y-a.a.y)/(a.b.x-a.a.x));double b2=b.a.y-b.a.x*((b.b.y-b.a.y)/(b.b.x-b.a.x));if(b1==b2)cout<<"LINE"<<endl;elsecout<<"NONE"<<endl;}else                  //相交 {double a1,a2,b1,b2,c1,c2,x,y;a1=a.b.x-a.a.x;a2=b.b.x-b.a.x;b1=a.a.y-a.b.y;b2=b.a.y-b.b.y;c1=a.a.x*a.b.y-a.a.y*a.b.x;c2=b.a.x*b.b.y-b.a.y*b.b.x;x=(c1*a2-c2*a1)/(a1*b2-a2*b1);            y=(c1*b2-c2*b1)/(a2*b1-a1*b2);            printf("POINT %.2lf %.2lf\n",x,y);}}cout<<"END OF OUTPUT"<<endl;return 0;}