POJ 1269 Intersecting Lines

判断直线相交及求交点，可能会有斜率不存在的情况

#include <iostream>#include <cstdio>int main(){    double x1, y1, x2, y2, x3, y3, x4, y4;    int n;    double k1, k2;    double b1, b2;    double i_x, i_y;    scanf("%d", &n);    std::cout << "INTERSECTING LINES OUTPUT" << std::endl;    while( n-- )    {        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4);        if( x1 != x2 && x3 != x4 )        {            k1 = ( y2 - y1 ) / ( x2 - x1 );            k2 = ( y4 - y3 ) / ( x4 - x3 );            b1 = y1 - k1 * x1;            b2 = y3 - k2 * x3;            if( k1 == k2 )            {                if( b1 == b2 )                    printf("LINE\n");                else                    printf("NONE\n");            }            else            {                i_x = (b2 - b1) / (k1 - k2);                i_y = k1 * i_x + b1;                printf("POINT %.2lf %.2lf\n", i_x, i_y);            }        }        else if( x1 == x2 && x3 == x4 )        {            if( x1 == x3 )            std::cout << "LINE\n";            else            std::cout << "NONE\n";        }        else if( x1 == x2 && x3 != x4 )        {            k2 = ( y4 - y3 ) / ( x4 - x3 );            b2 = y3 - k2 * x3;            i_x = x1;            i_y = k2 * x1 + b2;            printf("POINT %.2lf %.2lf\n", i_x, i_y);        }        else        {            k1 = ( y2 - y1 ) / ( x2 - x1 );            b1 = y1 - k1 * x1;            i_x = x3;            i_y = k1 * x3 + b1;            printf("POINT %.2lf %.2lf\n", i_x, i_y);        }    }    std::cout << "END OF OUTPUT\n";    return 0;}