【BZOJ1226】学校食堂Dining（状压DP）

题意：见题面

思路：设dp[i,sta,k]为前i个人已经吃完，从第i人到第i+b[i]人的吃饭状况是sta，前一个吃完的人离i的距离是k（可能为负）的最小值

\[ dp[i+1,sta>>1,k-1]=min(dp[i+1,sta>>1,k-1],dp[i,sta,k]) (sta and 1=1，如果i已经吃完)\]

\[dp[i,sta+1<<l,l]=min(dp[i,sta+1<<l,l],dp[i,sta,k]+w(i+k,i+l) (第i+l人吃，前一个吃的人是i+k)\]

\[w(i,j)=b[i]  or  b[j]-b[i]  and  b[j]=b[i]  xor  b[j]\]

注意转移的时候边界判断，是否超过l人里面最小的容忍范围

 const oo=;
 var dp:array[..,..,-..]of longint;
     t,b:array[..]of longint;
     cas,v,n,i,j,k,l,ans,r:longint;

 function clac(x,y:longint):longint;
 begin
  if x= then exit();
  exit(t[x] xor t[y]);
 end;

 function min(x,y:longint):longint;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 begin
  assign(input,'bzoj1226.in'); reset(input);
  assign(output,'bzoj1226.out'); rewrite(output);
  readln(cas);
  for v:= to cas do
  begin
   read(n);
   for i:= to n do read(t[i],b[i]);
   fillchar(dp,sizeof(dp),$7f);
   dp[,,-]:=;
   for i:= to n do
    for j:= to (<<)- do
     for k:=- to  do
      if dp[i,j,k]<oo then
      begin
       if j and > then
        dp[i+,j>>,k-]:=min(dp[i+,j>>,k-],dp[i,j,k])
         else
         begin
          r:=oo;
          for l:= to  do
           if j and (<<l)= then
           begin
            if i+l>r then break;
            r:=min(r,i+b[i+l]+l);
            dp[i,j+(<<l),l]:=min(dp[i,j+(<<l),l],dp[i,j,k]+clac(i+k,i+l));
           end;
         end;
      end;
   ans:=oo;
   for i:=- to - do ans:=min(ans,dp[n+,,i]);
   writeln(ans);
  end;
  close(input);
  close(output);
 end.