Line belt(hdu 3400,内嵌三分)
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http://acm.hdu.edu.cn/showproblem.php?pid=3400
Line belt
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2245 Accepted Submission(s): 850
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
解析:
题意:
给出A,B,C,D四个点,已知分别在AB,CD线段上行走速度以及在在这两条线段之外的行走速度,求从A点出发到达D点的最少时间是多少
思路:
内嵌三分,
对于坐标点进行三分
268 KB 0 ms C++ 1288 B
*/
#include<stdio.h>#include<math.h>#include<stdlib.h>#include<string.h>#include <iostream>using namespace std;const double eps=1e-9;double P,Q,R;struct point{double x;double y;};double len(point a1,point a2){return sqrt((a1.x-a2.x)*(a1.x-a2.x)+(a1.y-a2.y)*(a1.y-a2.y));}double solve1(point a,point b,point m){point mid,mmid,l,r;l=a;r=b;double t1,t2;do{mid.x=(l.x+r.x)/2.0;mid.y=(l.y+r.y)/2.0;mmid.x=(r.x+mid.x)/2.0;mmid.y=(r.y+mid.y)/2.0;t1=len(a,mid)/P+len(mid,m)/R;t2=len(a,mmid)/P+len(mmid,m)/R;if(t1<t2)r=mmid;elsel=mid;}while(len(r,l)>eps);return t1;}double solve2(point c,point d,point a,point b){point mid,mmid,l,r;l=d;r=c;double t1,t2;do{mid.x=(l.x+r.x)/2.0;mid.y=(l.y+r.y)/2.0;mmid.x=(r.x+mid.x)/2.0;mmid.y=(r.y+mid.y)/2.0;t1=len(d,mid)/Q+solve1(a,b,mid);t2=len(d,mmid)/Q+solve1(a,b,mmid);if(t1<t2)r=mmid;elsel=mid;}while(len(l,r)>eps);return t1;}int main(){ int T; scanf("%d",&T); while(T--) { point a,b,c,d; 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); double ans;ans=solve2(c,d,a,b); printf("%.2lf\n",ans); } //system("pause"); return 0;}
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