ACM--steps--4.1.6--Line belt

Line belt

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

Total Submission(s): 426 Accepted Submission(s): 228

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

Author lxhgww&&momodi

Source HDOJ Monthly Contest – 2010.05.01

Recommend lcy #include #include #include using namespace std; //我反正是没看出来为啥是凹性函数，不过要求一个最小时间，肯定会有一个最低点，，， double p,q,r;//表示三个速度。 struct point { double x,y;//坐标点； }; double dis(point a,point b) { //计算两点之间的距离 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } double dyx(point a,point c,point d) { double t1,t2;//表示两个时间。 //t1=t2=0; point left,right,mid,mmid; left=c; right=d; do { //对CD线段进行三分。 mid.x=(left.x+right.x)/2; mid.y=(left.y+right.y)/2; mmid.x=(mid.x+right.x)/2; mmid.y=(mid.y+right.y)/2; t1=dis(a,mid)/r+dis(mid,d)/q; t2=dis(a,mmid)/r+dis(mmid,d)/q; if(t1>t2) left=mid; else right=mmid; }while(fabs(t1-t2)>0.000001); return t1; } double wyx(point a,point b,point c,point d) { //先三分第一条线段，找出一个点。 double t1,t2;//表示两个时间。 t1=t2=0; point left,right,mid,mmid; left=a; right=b; do { //三分法先取两点之间的中点，再取中点与右端点的中点。 mid.x=(left.x+right.x)/2; mid.y=(left.y+right.y)/2; mmid.x=(mid.x+right.x)/2; mmid.y=(mid.y+right.y)/2; t1=dis(mid,a)/p+dyx(mid,c,d); t2=dis(mmid,a)/p+dyx(mmid,c,d); if(t1>t2) left=mid; else right=mmid; }while(fabs(t1-t2)>0.000001); return t1; } int main() { int T; cin>>T; point a,b,c,d; while(T--) { cin>>a.x>>a.y>>b.x>>b.y; cin>>c.x>>c.y>>d.x>>d.y; cin>>p>>q>>r; cout<