6Line belt
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简单题意
有两条线段AB,CD,走AB段速度是P,走CD段速度是Q,其他路段速度都是R,计算从A走到D的最短时间
解题思路形成过程
在AB上找一个点X,在CD上找一个点Y,按照A-X-Y-D的路线走,分别控制一边利用三分求出最短时间。
感想
真的没想到三分还能这样用,利用两层嵌套求解。
AC代码
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<string.h>
using namespace std;
struct point
{
double x,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 p,q,r;
double find2(point a,point c,point d)
{
point left,right;
point mid,midmid;
double t1,t2;
left=c;right=d;
do
{
mid.x=(left.x+right.x)/2;
mid.y=(left.y+right.y)/2;
midmid.x=(mid.x+right.x)/2;
midmid.y=(mid.y+right.y)/2;
t1=dis(a,mid)/r+dis(mid,d)/q;
t2=dis(a,midmid)/r+dis(midmid,d)/q;
if(t1>t2)left=mid;
else right=midmid;
}
while(dis(left,right)>=1e-5);
return t1;
}
double find(point a,point b,point c,point d)
{
point left,right;
point mid,midmid;
double t1,t2;
left=a;
right=b;
do
{
mid.x=(left.x+right.x)/2;
mid.y=(left.y+right.y)/2;
midmid.x=(mid.x+right.x)/2;
midmid.y=(mid.y+right.y)/2;
t1=dis(a,mid)/p+find2(mid,c,d);
t2=dis(a,midmid)/p+find2(midmid,c,d);
if(t1>t2)left=mid;
else right=midmid;
}while(dis(right,left)>=1e-5);
return t1;
}
int main()
{
freopen("in.txt","r",stdin);
int t;
point a,b,c,d;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&d.x,&d.y,&p,&q,&r);
printf("%.2lf\n",find(a,b,c,d));
}
return 0;
}
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