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。
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。
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)。
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
代码例如以下:
#include<cstdio>
#include<cmath>
double
dis(
double
x1,
double
y1,
double
x2,
double
y2)
{
return
sqrt
((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));
}
double
v1, v2, v3, x[4], y[4];
double
mindis(
double
xi,
double
yi)
{
double
i = x[3] - x[2];
double
j = y[3] - y[2];
double
l = 0, r = 1;
int
t;
double
tmid, tmidmid;
for
(t = 1; t <= 50; t++)
{
double
mid = (l + r) / 2;
double
midmid = (mid + r) / 2;
double
midx = x[2] + mid*i;
double
midy = y[2] + mid*j;
double
midmidx = x[2] + midmid*i;
double
midmidy = 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;
else
r = midmid;
}
return
tmidmid;
}
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);
double
i = x[1] - x[0];
double
j = y[1] - y[0];
double
l = 0, r = 1;
int
t;
double
tmid, tmidmid;
for
(t = 1; t <= 50; t++)
{
double
mid = (l + r) / 2;
double
midmid = (mid + r) / 2;
double
midx = x[0] + mid*i;
double
midy = y[0] + mid*j;
double
midmidx = x[0] + midmid*i;
double
midmidy = 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;
else
r = midmid;
}
printf
(
"%.2lf
"
, tmid);
}
return
0;
}