hdoj_3400Line belt

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1885    Accepted Submission(s): 713

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

10 0 0 100100 0 100 1002 2 1

 

Sample Output

136.60

 

三分嵌套三分。

AB上确定一点，然后三分枚举CD上的点。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const double EPS = 1e-10;int p,q,r;struct point {double x;double y;};double dis(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double findy(point c,point d,point y){point mid,midmid,left,right;double mid_t,midmid_t;left = c;right = d;do{mid.x = (left.x + right.x) / 2;mid.y = (left.y + right.y) / 2;midmid.x = (right.x + mid.x) / 2;midmid.y = (right.y + mid.y) / 2;mid_t = dis(d,mid) / q + dis(mid,y) / r;midmid_t = dis(d,midmid) / q+dis(midmid,y) / r;if(mid_t > midmid_t)left = mid;else right = midmid;}while(fabs(mid_t - midmid_t)>EPS);return mid_t;}double find(point a,point b,point c,point d){point mid,midmid,left,right;double mid_t,midmid_t;left = a;right = b;do{mid.x = (left.x + right.x) / 2;mid.y = (left.y + right.y) / 2;midmid.x = (right.x + mid.x) / 2;midmid.y = (right.y + mid.y) / 2;mid_t = dis(a,mid) / p + findy(c,d,mid);midmid_t = dis(a,midmid) / p + findy(c,d,midmid);if(mid_t > midmid_t)left = mid;else right = midmid;}while(fabs(mid_t - midmid_t)>EPS);return mid_t;}int main(){freopen("in.txt","r",stdin);int t;point a,b,c,d;cin>>t;while(t--){cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y>>d.x>>d.y;cin>>p>>q>>r;printf("%.2lf\n",find(a,b,c,d));}return 0;}