bzoj3205

和bzoj2595类似，也是斯坦纳树

设f[l,r,]表示在点i机器人组合成了l-r最少推的次数，然后可得

f[l,r,i]=min(f[l,m,i]+f[m+1,r,i])

f[l,r,i]=min(f[l,r,j]+1) 点j能推到i

但是这样做肯定会TLE，考虑两个优化

首先，一开始其实有很多根本用不到，我们可以先从机器人初始位置搜下去，找到所有可以访问的点做dp即可

其次，观察第二个方程，它的边权都是1，我们一定要用spfa转移吗？不，我们可以用直接宽搜

具体的我们维护两个队列，第一个队列是初始的按从小到大排，新加入的点放在第二个队列，每次取两个队头小的那一个

但我还是tle，求神犇指教

 const dx:array[..] of longint=(,,,-);
       dy:array[..] of longint=(,,-,);
       inf=;
 type node=record
        x,y:longint;
      end;

 var w:array[..] of node;
     q1,q2,st:array[..] of longint;
     po:array[..,..,..] of longint;
     f:array[..,..,..] of longint;
     c:array[..,..] of char;
     v:array[..] of boolean;
     loc:array[..] of longint;
     next:array[..,..] of longint;
     s:array[..] of longint;
     mid,h1,t1,i,j,p,q,l,x,y,k,n,m,ans,mx,tot:longint;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 function dfs(x,y,k:longint):longint;
   var xx,yy:longint;
   begin
     if (po[x,y,k]=) then exit(-);
     if po[x,y,k]> then exit(po[x,y,k]);
     po[x,y,k]:=;
     xx:=x+dx[k];
     yy:=y+dy[k];
     if (xx<) or (xx>n) or (yy<) or (yy>m) or (c[xx,yy]='x') then po[x,y,k]:=(x-)*m+y
     else if c[xx,yy]='A' then po[x,y,k]:=dfs(xx,yy,(k+) mod +)
     else if c[xx,yy]='C' then po[x,y,k]:=dfs(xx,yy,k mod +)
     else po[x,y,k]:=dfs(xx,yy,k);
     exit(po[x,y,k]);
   end;

 function cmp(i,j,l,r:longint):boolean;
   begin
     exit(f[i,l,r]<f[j,l,r]);
   end;

 procedure sort(l,r:longint);
   var i:longint;
   begin
     for i:= to mx do
       inc(s[i],s[i-]);
     for i:=t1 downto  do
     begin
       q1[s[f[st[i],l,r]]]:=st[i];
       dec(s[f[st[i],l,r]]);
     end;
   end;

 procedure bfs(l,r:longint);
   var h2,t2,i,x,y:longint;
   begin
     h2:=;
     t2:=;
     while (h2<=t2) or (h1<=t1) do
     begin
       if (h2>t2) or (h1<=t1) and cmp(q1[h1],q2[h2],l,r) then
       begin
         x:=q1[h1];
         inc(h1);
       end
       else begin
         x:=q2[h2];
         inc(h2);
       end;
       v[x]:=true;
       for i:= to  do
       begin
         if next[x,i]= then continue;
         y:=next[x,i];
         if not v[y] and (f[x,l,r]+<f[y,l,r]) then
         begin
           v[y]:=true;
           f[y,l,r]:=f[x,l,r]+;
           inc(t2);
           q2[t2]:=y;
         end;
       end;
     end;
   end;

 begin
   readln(k,m,n);
   for i:= to n*m do
     for p:= to k do
       for q:= to k do
         f[i,p,q]:=inf;
   for i:= to n do
   begin
     for j:= to m do
     begin
       read(c[i,j]);
       if (c[i,j]>='') and (c[i,j]<='') then
       begin
         x:=ord(c[i,j])-;
         inc(t1);
         w[t1].x:=i; w[t1].y:=j;
         f[t1,x,x]:=;
         loc[(i-)*m+j]:=t1;
       end;
     end;
     readln;
   end;
   fillchar(po,sizeof(po),);
   h1:=;
   while h1<=t1 do
   begin
     x:=w[h1].x; y:=w[h1].y;
     for i:= to  do
     begin
       if po[x,y,i]=- then
       begin
         po[x,y,i]:=dfs(x,y,i);
         if (po[x,y,i]>) and (loc[po[x,y,i]]=) then
         begin
           inc(t1);
           w[t1].x:=(po[x,y,i]-) div m+;
           w[t1].y:=(po[x,y,i]-) mod m+;
           loc[po[x,y,i]]:=t1;
         end;
       end;
       if po[x,y,i]> then next[h1,i]:=loc[po[x,y,i]];
     end;
     inc(h1);
   end;
   tot:=t1;
   for l:= to k do
     for p:= to k-l+ do
     begin
       q:=p+l-;
       h1:=; t1:=; mx:=;
       for i:= to tot do
       begin
         v[i]:=false;
         for mid:=p to q- do
           f[i,p,q]:=min(f[i,p,q],f[i,p,mid]+f[i,mid+,q]);
         if f[i,p,q]<inf then
         begin
           inc(t1);
           st[t1]:=i;
           inc(s[f[i,p,q]]);
           if f[i,p,q]>mx then mx:=f[i,p,q];
         end;
       end;
       sort(p,q);
       for i:= to mx do
         s[i]:=;
       bfs(p,q);
     end;
   ans:=inf;
   for i:= to tot do
     ans:=min(ans,f[i,,k]);
   if ans>=inf then writeln(-)
   else writeln(ans);
 end.