題意:平面內有兩條線段，在兩條線段上的速度v1，v2，其他地方速度爲v0，求從一條線段起點到另一條線段的終點的最短時間。
題解:從一個點出發到另一條線段的終點的時間是凹函數(證明略。三分套三分。第一次三分第一條線段上的點，第二次三分出經過這個點的ans最小值。
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define eps 1e-3
double xa,xb,xc,xd,ya,yb,yc,yd;
double v1,v2,v0;
double dis(double x1,double y1,double x2,double y2){
return (double)sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double t(double x1,double y1,double x2,double y2){
double ans=0;
ans+=(double)(dis(x1,y1,xa,ya)/v1);
ans+=(double)(dis(x1,y1,x2,y2)/v0);
ans+=(double)(dis(x2,y2,xd,yd)/v2);
return (double)ans;
}
double calc(double x,double y){
double lx=xc,ly=yc,rx=xd,ry=yd;
while(dis(lx,ly,rx,ry)>eps){
double deltax=(rx-lx)/3;
double deltay=(ry-ly)/3;
double midx=lx+deltax;
double midy=ly+deltay;
double mmx=midx+deltax;
double mmy=midy+deltay;
if(t(x,y,midx,midy)<t(x,y,mmx,mmy)) rx=mmx,ry=mmy;
else lx=midx,ly=midy;
}
return (double)t(x,y,lx,ly);
}
int main(){
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
scanf("%lf%lf%lf%lf",&xa,&ya,&xb,&yb);
scanf("%lf%lf%lf%lf",&xc,&yc,&xd,&yd);
scanf("%lf%lf%lf",&v1,&v2,&v0);
double lx=xa,ly=ya,rx=xb,ry=yb;
while(dis(lx,ly,rx,ry)>eps){
double deltax=(rx-lx)/3;
double deltay=(ry-ly)/3;
double midx=lx+deltax;
double midy=ly+deltay;
double mmx=midx+deltax;
double mmy=midy+deltay;
if(calc(midx,midy)<calc(mmx,mmy)) rx=mmx,ry=mmy;
else lx=midx,ly=midy;
}
printf("%.2lf",calc(lx,ly));
return 0;
}