bzoj2165

类似于bzoj1706，考虑到楼层是等价的
我们令f[p,i,j]为用了2^p次电梯，从房间i到j最多上升了多少层
然后从2^p很容易推到2^(p+1)，类似于floyd的转移即可
下面我们要用最小的电梯次数
可以考虑每一个数都有其唯一的而二进制拆分
从p到0贪心得到一个最接近上了m层的次数ans,ans+1即为答案

 const inf=-;
 var f:array[..,..,..] of int64;
     w,v:array[..] of int64;
     e,t,i,j,k,n,p,h:longint;
     fl:boolean;
     ans,m:int64;

 function max(a,b:int64):int64;
   begin
     if a>b then exit(a) else exit(b);
   end;

 function check(p:longint):boolean;
   var i:longint;
   begin
     for i:= to n do
       if f[p,,i]>=m then exit(true);
     exit(false);
   end;

 begin
   readln(t);
   for e:= to t do
   begin
     readln(n,m);
     for i:= to n do
     begin
       for j:= to n do
       begin
         read(f[,i,j]);
         if f[,i,j]= then f[,i,j]:=inf;
       end;
       readln;
     end;
     p:=;
     while true do
     begin
       for i:= to n do
         for j:= to n do
         begin
           f[p+,i,j]:=inf;
           for k:= to n do
           begin
             f[p+,i,j]:=max(f[p+,i,j],f[p,i,k]+f[p,k,j]);
             if f[p+,i,j]>m then
             begin
               f[p+,i,j]:=m;
               break;
             end;
           end;
         end;
       if not check(p+) then inc(p)  //出现了从房间1到某房间最大上升层数大于m就不用往下穷举了
       else break;
     end;
     w[]:=;
     for i:= to n do
       w[i]:=inf;  //w表示在当前所用次数下到i号房间最大上升层数层数
     ans:=;
     for h:=p downto  do
     begin
       fl:=true;
       for i:= to n do
       begin
         for j:= to n do
           if w[i]+f[h,i,j]>=m then
           begin
             fl:=false;
             break;
           end;
         if not fl then break;
       end;
       if fl then
       begin
         ans:=ans+int64() shl h;
         for i:= to n do
           v[i]:=inf;
         for i:= to n do
           for j:= to n do
             v[j]:=max(v[j],w[i]+f[h,i,j]);
         w:=v;
       end;
     end;
     writeln(ans+);
   end;
 end.