2013 - ECJTU 暑期训练赛第七场-problem-A

A -O.O

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:10000KB    64bit IO Format:%I64d & %I64u

SubmitStatusPractice POJ 1269

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

AC代码+解释：

#include<iostream>#include<cstdio>#include<iomanip>using namespace std;double x1,x2,x3,x4,y1,y2,y3,y4,k1,k2,b1,b2,x,y;int main(){    int n,i;    while(cin>>n)    {        cout<<"INTERSECTING LINES OUTPUT"<<endl;        for(i=0; i<n; i++)        {            cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;            if(x1!=x2&&x3!=x4)            {                k1=(y1-y2)/(x1-x2);//直线A->B的斜率                k2=(y3-y4)/(x3-x4);//直线C->D的斜率                b1=y1-(k1*x1);//直线A->B的b                b2=y3-(k2*x3);//直线C->D的b                if(k1==k2)//如果相等，表明不是平行就是重合                {                    if((x3*k1+b1)==y3)//如果直线C->D上点在直线A->B上表明两条直线重合了                        cout<<"LINE"<<endl;                    else//否则平行                        cout<<"NONE"<<endl;                }                if(k1!=k2)//如果斜率不相等，表明斜率一定存在                {                    x=(b2-b1)/(k1-k2);                    y=(k1*x)+b1;                    cout<<"POINT ";                    printf("%.2lf %.2lf\n",x,y);                }            }            if(x1==x2||x3==x4)//考虑两条直线的斜率都不存在的情况            {                if(x1==x2&&x3==x4)//如果直线A->B和C->D的两个点横坐标相等                {                    if(x1==x3)//如果四个点都在同一条直线上                        cout<<"LINE"<<endl;                    else//否则就平行                        cout<<"NONE"<<endl;                }                if(x1==x2&&x3!=x4)//下面两种情况是如果有一条直线斜率不存在而另一条直线存在                {                    k2=(y3-y4)/(x3-x4);                    b2=y3-k2*x3;                    y1=k2*x1+b2;                    cout<<"POINT ";                    printf("%.2lf %.2lf\n",x1,y1);                }                if(x1!=x2&&x3==x4)                {                    k1=(y1-y2)/(x1-x2);                    b1=y1-k1*x1;                    y3=k1*x3+b1;                    cout<<"POINT ";                    printf("%.2lf %.2lf\n",x3,y3);                }            }        }        cout<<"END OF OUTPUT"<<endl;    }    return 0;}