Asakura的魔法世界

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Description

Asakura存在于一个魔法世界中。
有一天，Asakura在一条魔法通道里偷懒，突然接到一个紧急任务，要高速赶往还有一条通道b去。
我们把通道a和b看作两条线段AB和CD。Asakura初始位置在A，如今要高速赶往D。
Asakura在魔法通道a上的速度为v1。在魔法通道b上速度为v2，在除了这两条通道上的其余位置的速度为v3。Asakura最快多长时间才干到达指定位置。

Input

多组输入
对于每组測试数据：包括三行
第一行输入四个整数x1,y1,x2,y2.当中(x1,y1)为A。(x2,y2)为B.(-10000<=x1,y1,x2,y2<=10000)
第二行输入四个整数x3,y3,x4,y4.当中(x3,y3)为C，(x4,y4)为D.(-10000<=x3,y3,x4,y4<=10000)
第三行输入三个整数v1,v2,v3.(1<=v1,v2,v3<=50)
每两组測试数据间输入一个空行。

Output

输出A到D最短的时间。结果精度到小数点后两位（四舍五入）。

Sample Input

0 0 50 50
50 0 50 50
1 1 1

0 0 50 50
50 0 50 50
3 1 3

0 0 50 50
50 0 50 50
1 3 1

Sample Output

70.71
23.57
63.81







代码例如以下：

#include<cstdio>

#include<cmath>

doubledis(double x1, double y1,double x2, double y2)

{

    returnsqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));

}

doublev1, v2, v3, x[4], y[4];

doublemindis(doublexi, double yi)

{

    doublei = x[3] - x[2];

    doublej = y[3] - y[2];

    doublel = 0, r = 1;

    intt;

    doubletmid, tmidmid;

    for(t = 1; t <= 50; t++)

    {

        doublemid = (l + r) / 2;

        doublemidmid = (mid + r) / 2;

        doublemidx = x[2] + mid*i;

        doublemidy = y[2] + mid*j;

        doublemidmidx = x[2] + midmid*i;

        doublemidmidy = y[2] + midmid*j;

        tmid = dis(xi, yi, midx, midy) / v3 + dis(midx, midy, x[3], y[3]) / v2;

        tmidmid = dis(xi, yi, midmidx, midmidy) / v3 + dis(midmidx, midmidy, x[3], y[3]) / v2;

        if(tmid > tmidmid)

            l = mid;

        elser = midmid;

    }

    returntmidmid;

}

int main()

{

    while(~scanf("%lf %lf %lf %lf", &x[0], &y[0], &x[1], &y[1]))

    {

        scanf("%lf %lf %lf %lf", &x[2], &y[2], &x[3], &y[3]);

        scanf("%lf %lf %lf", &v1, &v2, &v3);

        doublei = x[1] - x[0];

        doublej = y[1] - y[0];

        doublel = 0, r = 1;

        intt;

        doubletmid, tmidmid;

        for(t = 1; t <= 50; t++)

        {

            doublemid = (l + r) / 2;

            doublemidmid = (mid + r) / 2;

            doublemidx = x[0] + mid*i;

            doublemidy = y[0] + mid*j;

            doublemidmidx = x[0] + midmid*i;

            doublemidmidy = y[0] + midmid*j;

            tmid = dis(x[0], y[0], midx, midy) / v1 + mindis(midx, midy);

            tmidmid = dis(x[0], y[0], midmidx, midmidy) / v1 + mindis(midmidx, midmidy);

            if(tmid > tmidmid)

                l = mid;

            elser = midmid;

        }

        printf("%.2lf\n", tmid);

    }

    return0;

}