Three points of quality in hdu-3400

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Well, there is a quality three-point question, which belongs to the nested three-point question.

After thinking about this question for a long time, I only wrote several independent equations. Let's look at the solution report. Then, I began to think about how to use the three-way method. I have never been able to figure it out (under review, there is no independent analysis ).

Look at Daniel's problem-solving report to find out how to do it;

Well written. Http://hi.baidu.com/1093782566/blog/item/2bb8a017124a3735dd540126.html

Here we will mainly explain how three points are made.

Total time consumed T = x/P + Y/R + z/Q;
X = SQRT (XX-Ax) ^ 2 + (xy-ay) ^ 2), monotonic incrementing Function
Y = SQRT (-Yx + dx) ^ 2 + (dy-yy) ^ 2), monotonic decreasing function
Z = SQRT (Yx-xx) ^ 2 + (yy-XY) ^ 2) convex function.

Because Z is determined by X and Y, Z is increased first and then subtracted. First 3 X, then 3 Y. X is related to X (A, B) and Y (c, d. That is, you can nest the three-point search y based on the three-point search X.

Code:

#include<iostream>#include <math.h>#include <stdio.h>using namespace std;#define INF 0.00001class point{public:double x;double y;};point A,B,C,D;double P,R,Q;double cal_dis(point a,point b){return sqrt((b.y-a.y)*(b.y-a.y)+(b.x-a.x)*(b.x-a.x));}double cal_y(point a){point first=C;point end=D;point mid,midmid;double t1,t2;do{mid.x=(first.x+end.x)/2.0;mid.y=(first.y+end.y)/2.0;midmid.x=(mid.x+end.x)/2.0;midmid.y=(mid.y+end.y)/2.0;t1=cal_dis(mid,D)/Q+cal_dis(a,mid)/R;t2=cal_dis(midmid,D)/Q+cal_dis(a,midmid)/R;if(t1>t2) first=mid;else end=midmid;}while(fabs(t1-t2)>INF);return t1;}double cal_t(){point first=A;point end=B;point mid,midmid;double t1,t2;do{mid.x=(first.x+end.x)/2.0;mid.y=(first.y+end.y)/2.0;midmid.x=(mid.x+end.x)/2.0;midmid.y=(mid.y+end.y)/2.0;t1=cal_dis(mid,A)/P+cal_y(mid);t2=cal_dis(midmid,A)/P+cal_y(midmid);if(t1>t2) first=mid;else end=midmid;}while(fabs(t1-t2)>INF);return t1; }int main(){int t;cin>>t;while(t--){cin>>A.x>>A.y>>B.x>>B.y>>C.x>>C.y>>D.x>>D.y>>P>>Q>>R;printf("%.2lf\n",cal_t());}return 0;}

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