hdu 3400 Line belt 三分

Line belt

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3930 Accepted Submission(s): 1528



Problem Description

In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane.

How long must he take to travel from A to D?

Input

The first line is the case number T.

For each case, there are three lines.

The first line, four integers, the coordinates of A and B: Ax Ay Bx By.

The second line , four integers, the coordinates of C and D:Cx Cy Dx Dy.

The third line, three integers, P Q R.

0<= Ax，Ay，Bx，By，Cx，Cy，Dx，Dy<=1000

1<=P，Q，R<=10

Output

The minimum time to travel from A to D, round to two decimals.

Sample Input

1
0 0 0 100
100 0 100 100
2 2 1

Sample Output

136.60

题意：有一个线段AB，一个线段CD，有一个点从A出发，要走到D，在AB上移动速度是P，CD移动速度是Q，其他地方移动速度是R。问最短时间。

解法：三分，对线段AB进行三分，AB三分的两个点m=(l+r)/2和mm=(m+r)/2，两个点对应的最小时间lm，rm。对于m，计算lm = dis（A，m），再三分CD计算出CD某个点到m和D的时间最小值，返回给m，m点处的lm=dis(A,m)加上返回的这个值就是从A走到m点走到CD线段再走到D点的最小时间。对mm坐相同处理，值为rm。然后用求得lm和rm继续对AB进行三分求得最小时间。

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;

struct node{
double x,y;
}A,B,C,D;
double P,Q,R;

double dis(node a,node b) ///两点间距离
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double Fd(node c,node d,node mid)  ///三分CD线段到AB某个点的最短时间
{
node l = c,r = d,m,mm;
double lm = 1,rm = 2;
while(fabs(lm-rm) > eps){
m.x = (l.x+r.x)/2;
m.y = (l.y+r.y)/2;
mm.x = (m.x+r.x)/2;
mm.y = (m.y+r.y)/2;
lm = dis(d,m)/Q+dis(m,mid)/R;   ///CD线段走的时间和其他地方走的时间
rm = dis(d,mm)/Q+dis(mm,mid)/R;
if(lm > rm) l = m;
else        r = mm;
}
return lm;
}

int main(void)
{
int T;
scanf("%d",&T);
while(T--){
scanf("%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y);
scanf("%lf%lf%lf%lf",&C.x,&C.y,&D.x,&D.y);
scanf("%lf%lf%lf",&P,&Q,&R);
node l=A,r=B,m,mm;
double lm=10000,rm=20000; ///随便赋个不相同的值就行了
while(fabs(lm-rm) > eps){
m.x = (l.x+r.x)/2;
m.y = (l.y+r.y)/2;
mm.x = (m.x+r.x)/2;
mm.y = (m.y+r.y)/2;
lm = dis(A,m)/P+Fd(C,D,m);
rm = dis(A,mm)/P+Fd(C,D,mm);
if(lm > rm) l = m;
else        r = mm;
}
printf("%.2f\n",lm);
}
return 0;
}