三分套三分。
如果只有一条线段和一个点的话显然三分。。两条线段的话大概就是三分套三分了吧（反正我不会证
实测eps取1e-3就行了。。
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<cmath> 6 #define d double 7 #define y1 y11 8 using namespace std; 9 const d eps=1e-3; 10 int i,j,k,n,m,a,b,c; 11 d x1,x2,y11,y2,lx,ly,rx,ry,ans; 12 int sx1,sx2,sy1,sy2,tx1,tx2,ty1,ty2; 13 14 inline d dis(d x1,d y1,d x2,d y2){return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));} 15 inline d sum(d x1,d y1,d x2,d y2){ 16 return dis((d)sx1,(d)sy1,x1,y1)/(d)a+dis(x1,y1,x2,y2)/(d)c+dis(x2,y2,(d)tx2,(d)ty2)/(d)b; 17 } 18 d get(d X,d Y){ 19 d x1=tx1,x2=tx2,y1=ty1,y2=ty2,lx=x1,ly=y1,rx,ry,tim1,tim2,now=1002333333.0; 20 while(dis(x1,y1,x2,y2)>=eps){ 21 lx=x1+(x2-x1)/3.0,ly=y1+(y2-y1)/3.0; 22 rx=lx+lx-x1,ry=ly+ly-y1; 23 tim1=sum(X,Y,lx,ly),tim2=sum(X,Y,rx,ry); 24 25 if(tim1<tim2)x2=rx,y2=ry,now=min(now,tim1); 26 else x1=lx,y1=ly,now=min(now,tim2); 27 } 28 now=min(now,sum(X,Y,lx,ly)); 29 ans=min(ans,now); 30 return now; 31 } 32 33 int main(){ 34 scanf("%d%d%d%d",&sx1,&sy1,&sx2,&sy2); 35 scanf("%d%d%d%d",&tx1,&ty1,&tx2,&ty2); 36 scanf("%d%d%d",&a,&b,&c); 37 ans=1002333333.0; 38 39 x1=sx1,x2=sx2,y1=sy1,y2=sy2; 40 lx=x1,ly=y1; 41 while(dis(x1,y1,x2,y2)>=eps){ 42 lx=x1+(x2-x1)/3.0,ly=y1+(y2-y1)/3.0; 43 rx=lx+lx-x1,ry=ly+ly-y1; 44 if(get(lx,ly)<get(rx,ry))x2=rx,y2=ry; 45 else x1=lx,y1=ly; 46 } 47 ans=min(ans,get(x1,y1)); 48 printf("%.2lf ",ans); 49 return 0; 50 }
（反正怎么写都不会TLE，所以具体处理写得有点龊