2328: [HNOI2011]赛车游戏 - BZOJ

先把一定要减的费用先减掉，就是b*s*len（上坡路），下坡路就设一个初速度，使耗油为0，如果没油了，就无法到达

然后考虑加速

对于长度为len的路，增加v的速度需要len*a*v的油，与len成正比

然后又因为原来的速度越小，增加等量v的增益越高，所以每次都从最小的速度开始加（不超过第二小），还有不要超过vmax

 const
         maxn=;
 var
         t,n:longint;
         a,b,vmax,f:double;
         len,s,v:array[..maxn]of double;

 procedure init;
 var
         i:longint;
         x,y:double;
 begin
         read(a,b,vmax,f,n);
         for i:= to n do
           begin
             read(x,y);
             len[i]:=sqrt(sqr(x)+sqr(y))/;
             s[i]:=y/x;
             if s[i]> then f:=f-len[i]*b*s[i];
             if s[i]< then v[i]:=-b*s[i]/a
             else v[i]:=;
           end;
 end;

 procedure swap(var x,y:double);
 var
         t:double;
 begin
         t:=x;x:=y;y:=t;
 end;

 procedure sort(l,r:longint);
 var
         i,j:longint;
         y:double;
 begin
         i:=l;
         j:=r;
         y:=v[(l+r)>>];
         repeat
           while v[i]<y do
             inc(i);
           while v[j]>y do
             dec(j);
           if i<=j then
           begin
             swap(s[i],s[j]);
             swap(v[i],v[j]);
             swap(len[i],len[j]);
             inc(i);
             dec(j);
           end;
         until i>j;
         if i<r then sort(i,r);
         if j>l then sort(l,j);
 end;

 procedure work;
 var
         i,j:longint;
         sum:double;
 begin
         sort(,n);
         for i:= to n do
           if v[i]>vmax then v[i]:=vmax;
         v[n+]:=vmax;
         sum:=;
         for i:= to n do
           begin
             sum:=sum+len[i];
             if v[i]<>v[i+] then
             begin
               if f>=(v[i+]-v[i])*sum*a then
                 begin
                   f:=f-(v[i+]-v[i])*sum*a;
                   v[i]:=v[i+];
                 end
               else
                 begin
                   v[i]:=v[i]+f/(sum*a);
                   break;
                 end;
             end;
           end;
         for j:=i- downto  do
           v[j]:=v[j+];
         if v[]= then
         begin
           writeln('IMPOSSIBLE');
           exit;
         end;
         sum:=;
         for i:= to n do
           sum:=sum+len[i]/v[i];
         writeln(sum::);
 end;

 begin
         read(t);
         while t> do
           begin
             init;
             if f< then writeln('IMPOSSIBLE')
             else work;
             dec(t);
           end;
 end.