Bzoj1857:[Scoi2010]傳送帶:三分
阿新 • • 發佈:2019-01-23
首先由猜測法證得函式具有下凸性QwQ
然後就可以三分辣
#include<cmath> #include<cstdio> #include<cstdlib> #include<iostream> #include<algorithm> using namespace std; const double eps=1e-5; struct point{double x,y;}A,B,C,D; double p,q,r,ax,ay,bx,by,cx,cy,dx,dy; double powe(double x){return x*x;} double calc(point a,point b){ return sqrt(powe(a.x-b.x)+powe(a.y-b.y)); } double calc2(double x,double y,point a){ return sqrt(powe(x-a.x)+powe(y-a.y)); } double getval(point tmp){ C.x=cx; C.y=cy; D.x=dx; D.y=dy; while (abs(C.x-D.x)>eps||abs(C.y-D.y)>eps){ point tmpl; tmpl.x=C.x+(D.x-C.x)/3; tmpl.y=C.y+(-C.y+D.y)/3; point tmpr; tmpr.x=C.x+(D.x-C.x)/3*2; tmpr.y=C.y+(-C.y+D.y)/3*2; double ret1=calc2(ax,ay,tmp)/p+calc(tmp,tmpl)/r+calc2(dx,dy,tmpl)/q; double ret2=calc2(ax,ay,tmp)/p+calc(tmp,tmpr)/r+calc2(dx,dy,tmpr)/q; if (ret1>ret2) C=tmpl; else D=tmpr; } return calc2(ax,ay,tmp)/p+calc(tmp,C)/r+calc2(dx,dy,C)/q; } int main(){ 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); ax=A.x; ay=A.y; bx=B.x; by=B.y; cx=C.x; cy=C.y; dx=D.x; dy=D.y; scanf("%lf%lf%lf",&p,&q,&r); while (abs(A.x-B.x)>eps||abs(A.y-B.y)>eps){ point tmpl; tmpl.x=A.x+(B.x-A.x)/3; tmpl.y=A.y+(B.y-A.y)/3; point tmpr; tmpr.x=A.x+(B.x-A.x)/3*2; tmpr.y=A.y+(B.y-A.y)/3*2; double ret1=getval(tmpl),ret2=getval(tmpr); if(ret1>ret2) A=tmpl; else B=tmpr; } printf("%.2lf\n",getval(A)); }