poj 1269 Intersecting Lines(计算几何)

Intersecting Lines

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12284 Accepted: 5495

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

题意：有N组数据，每组数据分别给出两条直线的两个点，问这两条直线是否相交，不相交的话，如果共线则输出LINE，平行则输出NONE，相交的话，输出交点坐标

题解：三点共线的条件是：p1，p2，p3：(p2-p1)X(p3-p1)=0，没有交点就判断是否平行

#include<cstring>#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#define eps 1e-8using namespace std;struct point {    double x,y;} p[8];///叉积double ok(point a,point b,point c) {    return (c.x-a.x)*(b.y-a.y)-(b.x-a.x)*(c.y-a.y);}//共线判断bool Line(point a,point b,point c) {    return ok(a,b,c)==0;}///交点判断bool None() {    return (p[1].y-p[2].y)*(p[3].x-p[4].x)==(p[1].x-p[2].x)*(p[3].y-p[4].y);}//求交点point inter(point u1,point u2,point v1,point v2) {    point ret=u1;    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))             /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));    ret.x+=(u2.x-u1.x)*t;    ret.y+=(u2.y-u1.y)*t;    return ret;}int main() {    //freopen("in.txt","r",stdin);    int n;    while(cin>>n) {        printf("INTERSECTING LINES OUTPUT\n");        while(n--) {            for(int i=1; i<=4; i++)                scanf("%lf%lff",&p[i].x,&p[i].y);            if(Line(p[1],p[3],p[4])&&Line(p[2],p[3],p[4])) {                printf("LINE\n");                continue;            }            if(None()) {                printf("NONE\n");                continue;            }            point ans=inter(p[1],p[2],p[3],p[4]);            printf("POINT %.2f %.2f\n",ans.x,ans.y);        }        printf("END OF OUTPUT\n");    }    return 0;}