HDU/HDOJ 3400 Line belt 三分嵌套 杭电月赛

Line belt

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 24 Accepted Submission(s) : 6

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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

这个题思想很简单，就是三分重叠着用就是了

首先在AB上三分一个点出来，然后再在CD上三分一个点让值最小

两次三分重叠起来就可以解决题目了。。

先开始我的程序WA了好多次。一直不知道原因

后来查了下解题报告，发现他们说要再计算距离里面去加上一个精度

结果，果然就AC了。此题数据太YD了。。

另外，我三分的是距离A的距离和距离D的距离

我的代码：

#include<stdio.h>
#include<math.h>
#define eps 1e-9

struct point
{
double x;
double y;
};
point A,B,C,D,M1,M2;
double P,Q,R;

double dis(point a,point b)
{
return sqrt(eps+(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double cal2(double len)
{
double d1,d2,k,t1,t2;
d1=len,d2=dis(C,D);
k=d1/d2;
M2.x=(C.x-D.x)*k+D.x;
M2.y=(C.y-D.y)*k+D.y;
t1=dis(M1,M2)/R;
t2=len/Q;
return (t1+t2);
}

double cal1(double len)
{
int i;
double d1,d2,k,t1,tx,ty;
d1=len,d2=dis(A,B);
k=d1/d2;
M1.x=(B.x-A.x)*k+A.x;
M1.y=(B.y-A.y)*k+A.y;
t1=len/P;
double left,right,mid1,mid2;
left=0,right=dis(C,D);
for(i=1;i<=100;i++)
{
mid1=(2*left+right)/3;
mid2=(left+2*right)/3;
tx=cal2(mid1);
ty=cal2(mid2);
if(tx>ty)
{
left=mid1;
}
else
{
right=mid2;
}
}
return t1+cal2(left);
}

void triple()
{
int i;
double mid1,mid2,left,right,t1,t2;
left=0,right=dis(A,B);
for(i=1;i<=100;i++)
{
mid1=(left*2+right)/3;
mid2=(left+2*right)/3;
t1=cal1(mid1);
t2=cal1(mid2);
if(t1>t2)
{
left=mid1;
}
else
{
right=mid2;
}
}
printf("%.2lf\n",cal1(left));
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
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);
triple();
}
return 0;
}