poj 1269 Intersecting Lines (判断两条直线的状态）

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题目连接：http://poj.org/problem?id=1269

题目思路：给出两条直线，给出的方式是两个点。判断两条直线是平行，重合还是相交，相交的话求出交点。

解题思路：确定两点求直线，根据直线的参数去判断状态，要注意一些特殊情况，比如平行x,y的直线等。

#include <stdio.h>#include <string.h>double a, b, c, d;double x[4], y[4];double xi, yi;int main() {    int n;    while (scanf("%d", &n) == 1) {printf("INTERSECTING LINES OUTPUT\n");while (n--) {    for (int i = 0; i < 4; i++)scanf("%lf%lf", &x[i], &y[i]);    a = y[1] - y[0];    b = x[1] - x[0];    c = y[3] - y[2];    d = x[3] - x[2];    if (a * d == b * c && (d || a) && (b || c)) {if (b == 0 && d == 0) {    if (x[0] == x[2])printf("LINE\n");    elseprintf("NONE\n");}else if (y[0] - a * x[0] / b == y[2] - c * x[2] / d)    printf("LINE\n");else    printf("NONE\n");    }    else {if (b == 0) {    xi = x[0];    yi = c * (xi - x[2]) / d + y[2];}else if (d == 0) {    xi = x[2];    yi = a * (xi - x[0]) / b + y[0];}else {    xi = (b * d * (y[2] - y[0]) + a * d * x[0] - b * c * x[2]) / (a * d - b * c);    yi = a * (xi - x[0]) / b + y[0];}printf("POINT %.2lf %.2lf\n", xi, yi);    }}printf("END OF OUTPUT\n");    }    return 0;}