• POJ 1410 Intersection


    POJ_1410

        几个需要注意的地方:

        ①给出的矩形顶点的坐标需要自己重新排下序再使用,看discuss说所谓的top等并不是严格指上方,只是一个相对的参考,所以要重新排下序再用。

        ②因为题目里面说了矩形的内部也算矩形的一部分,所以线段在矩形内部是认为和矩形相交的。

        ③在判断线段与矩形四个边是否有交点时,要注意对非规范相交的判定,当线段和边共线且不相交时叉积也为0。

    #include<stdio.h>
    #include<string.h>
    #define zero 1e-8
    double xs, ys, xe, ye, xl, yl, xr, yr, x1[5], y1[5], x2[5], y2[5];
    double det(double x1, double y1, double x2, double y2)
    {
    return x1 * y2 - y1 * x2;
    }
    double fabs(double x)
    {
    return x < 0 ? -x : x;
    }
    double min(double x, double y)
    {
    return x < y ? x : y;
    }
    double max(double x, double y)
    {
    return x > y ? x : y;
    }
    int dcmp(double x)
    {
    if(fabs(x) < zero)
    return 0;
    if(x < 0)
    return -1;
    return 1;
    }
    int onseg(double x, double y, double x1, double y1, double x2, double y2)
    {
    if(fabs(x2 - x1) < fabs(y2 - y1))
    return dcmp((y - y2) * (y - y1)) <= 0;
    else
    return dcmp((x - x2) * (x - x1)) <= 0;
    }
    void solve()
    {
    int i, j, k;
    double t1, t2, t3, t4;
    x1[0] = xl, y1[0] = yr, x2[0] = xr, y2[0] = yr;
    x1[1] = xl, y1[1] = yr, x2[1] = xl, y2[1] = yl;
    x1[2] = xl, y1[2] = yl, x2[2] = xr, y2[2] = yl;
    x1[3] = xr, y1[3] = yr, x2[3] = xr, y2[3] = yl;
    for(i = 0; i < 4; i ++)
    {
    t1 = det(x2[i] - x1[i], y2[i] - y1[i], xs - x1[i], ys - y1[i]);
    t2 = det(x2[i] - x1[i], y2[i] - y1[i], xe - x1[i], ye - y1[i]);
    t3 = det(xe - xs, ye - ys, x1[i] - xs, y1[i] - ys);
    t4 = det(xe - xs, ye - ys, x2[i] - xs, y2[i] - ys);
    if(dcmp(t1) * dcmp(t2) < 0 && dcmp(t3) * dcmp(t4) < 0)
    {
    printf("T\n");
    return ;
    }
    if((dcmp(t1) == 0 && onseg(xs, ys, x1[i], y1[i], x2[i], y2[i])) || (dcmp(t2) == 0 && onseg(xe, ye, x1[i], y1[i], x2[i], y2[i])) || (dcmp(t3) == 0 && onseg(x1[i], y1[i], xs, ys, xe, ye)) ||(dcmp(t4) == 0 && onseg(x2[i], y2[i], xs, ys, xe, ye)))
    {
    printf("T\n");
    return ;
    }
    }
    printf("F\n");
    }
    int main()
    {
    int i, t;
    double x;
    scanf("%d", &t);
    while(t --)
    {
    scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &xs, &ys, &xe, &ye, &xl, &yl, &xr, &yr);
    if(xl > xr)
    x = xl, xl = xr, xr = x;
    if(yl < yr)
    x = yl, yl = yr, yr = x;
    if((xs < xr && xs > xl && ys < yl && ys > yr) || (xe < xr && xe > xl && ye < yl && ye > yr))
    printf("T\n");
    else
    solve();
    }
    return 0;
    }


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  • 原文地址:https://www.cnblogs.com/staginner/p/2346087.html
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