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?
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
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
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; }
Keep it simple!