2328: [HNOI2011]赛车游戏

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=10010;
var
        t,n:longint;
        a,b,vmax,f:double;
        len,s,v:array[0..maxn]of double;
 
procedure init;
var
        i:longint;
        x,y:double;
begin
        read(a,b,vmax,f,n);
        for i:=1 to n do
          begin
            read(x,y);
            len[i]:=sqrt(sqr(x)+sqr(y))/1000;
            s[i]:=y/x;
            if s[i]>0 then f:=f-len[i]*b*s[i];
            if s[i]<0 then v[i]:=-b*s[i]/a
            else v[i]:=0;
          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)>>1];
        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(1,n);
        for i:=1 to n do
          if v[i]>vmax then v[i]:=vmax;
        v[n+1]:=vmax;
        sum:=0;
        for i:=1 to n do
          begin
            sum:=sum+len[i];
            if v[i]<>v[i+1] then
            begin
              if f>=(v[i+1]-v[i])*sum*a then
                begin
                  f:=f-(v[i+1]-v[i])*sum*a;
                  v[i]:=v[i+1];
                end
              else
                begin
                  v[i]:=v[i]+f/(sum*a);
                  break;
                end;
            end;
          end;
        for j:=i-1 downto 1 do
          v[j]:=v[j+1];
        if v[1]=0 then
        begin
          writeln('IMPOSSIBLE');
          exit;
        end;
        sum:=0;
        for i:=1 to n do
          sum:=sum+len[i]/v[i];
        writeln(sum:0:5);
end;
 
begin
        read(t);
        while t>0 do
          begin
            init;
            if f<0 then writeln('IMPOSSIBLE')
            else work;
            dec(t);
          end;
end.