poj 1269 Intersecting Lines(判断直线相交 求交点)

Intersecting Lines

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6884 Accepted: 3207

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

题目：http://poj.org/problem?id=1269

题意：判断两条直线怎样相交，并输出可能的交点

分析：模板题，主要判断两条直线相同的时候，满足abs(a1*b2-a2*b1)<1e-8之后，

            要同时满足abs(a1*c2-a2*c1)<1e-8和abs(b1*c2-b2*c1)<1e-8才能说他们相同，否则不行，具体原因我觉得是精度问题吧= =

以后写代码都要写个test函数，里面测试各个函数是否正确，貌似是个好主意，减少失误率，方便查错 ^_^

代码：

#include<cmath>#include<cstdio>#include<iostream>using namespace std;typedef double mType;struct Tpoint{    mType x,y;    Tpoint(){}    Tpoint(mType _x,mType _y):x(_x),y(_y){}};struct Tline{    mType A,B,C;    Tline(){}    Tline(Tpoint s,Tpoint t)    {        A=s.y-t.y;        B=t.x-s.x;        C=s.x*t.y-s.y*t.x;    }    Tpoint IntersectPoint(Tline P)    {        mType tmp=P.B*A-P.A*B;        return Tpoint((P.C*B-P.B*C)/tmp,(P.A*C-P.C*A)/tmp);    }    int IsIntersect(Tline P)    {        if(fabs(A*P.B-B*P.A)<1e-8)        {            /**这里注意AC和BC都要判断,不然会WA*/            if(fabs(A*P.C-C*P.A)<1e-8&&fabs(B*P.C-C*P.B)<1e-8)return -1;            return 0;        }        return 1;    }};void test(){    Tline P=Tline(Tpoint(0,0),Tpoint(1,1));    Tline Q=Tline(Tpoint(1,0),Tpoint(2,1));    printf("%.2lf %.2lf %.2lf\n",P.A,P.B,P.C);    printf("%.2lf %.2lf %.2lf\n",Q.A,Q.B,Q.C);    if(P.IsIntersect(Q)>0)    {        puts("P is intersect with Q!");        Tpoint tmp=P.IntersectPoint(Q);        printf("The intersect point is:   %.2lf %.2lf\n",tmp.x,tmp.y);    }    else printf("no intersect find! state: %d\n",P.IsIntersect(Q));}int main(){    //test();    double sx,sy,tx,ty;    Tline P,Q;    Tpoint tmp;    int state,n;    scanf("%d",&n);    puts("INTERSECTING LINES OUTPUT");    while(n--)    {        scanf("%lf%lf%lf%lf",&sx,&sy,&tx,&ty);        P=Tline(Tpoint(sx,sy),Tpoint(tx,ty));        scanf("%lf%lf%lf%lf",&sx,&sy,&tx,&ty);        Q=Tline(Tpoint(sx,sy),Tpoint(tx,ty));        if((state=P.IsIntersect(Q))>0)        {            tmp=P.IntersectPoint(Q);            printf("POINT %.2lf %.2lf\n",tmp.x,tmp.y);        }        else printf(state<0?"LINE\n":"NONE\n");    }    puts("END OF OUTPUT");    return 0;}