poj3373

其实这道题只告诉了一个事
当出现多个满足答案约束条件是，我们可以求一个再求一个，不要一下子全求完
前两个条件怎么弄之前已经做过类似的了
于是我们可以用记忆化搜索找出最小差异
然后配合最小差异来剪枝，搜索出最小答案

 var a,b:array[..] of longint;
     f:array[..,..] of longint;
     n,m,k,i:longint;
     can:boolean;
     s:string;

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

 function differ(j,cur:longint):longint;  //f[j,cur]表示处理到第j位前面余数为cur的情况最小差异数
   var i,wh,s:longint;
   begin
     if (j=n+) then
     begin
       if cur= then exit()
       else exit();
     end;
     if f[j,cur]<>- then exit(f[j,cur]);
     if j<> then s:= else s:=;
     f[j,cur]:=;
     for i:=s to  do
     begin
       wh:=differ(j+,(cur*+i) mod m);
       if i<>a[j] then inc(wh);
       f[j,cur]:=min(f[j,cur],wh);
     end;
     exit(f[j,cur]);
   end;

 procedure dfs(j,wh,cur:longint);
   var i,s:longint;
   begin
     if (j=n+) then
     begin
       if cur= then can:=true;
       exit;
     end;
     if can then exit;
     if f[j,cur]+wh>k then exit;  //剪枝
     if j<> then s:= else s:=;
     for i:=s to  do
     begin
       b[j]:=i;
       if (i=a[j]) then
         dfs(j+,wh,(cur*+i) mod m)
       else
         dfs(j+,wh+,(cur*+i) mod m);
       if can then exit;
     end;
   end;

 begin
   while not eof do
   begin
     readln(s);
     n:=length(s);
     for i:= to n do
       a[i]:=ord(s[i])-;
     readln(m);
     if s='' then
     begin
       writeln();
       continue;
     end;
     fillchar(f,sizeof(f),);
     k:=differ(,);
 //    writeln(k);
     can:=false;
     dfs(,,);
     for i:= to n do
       write(b[i]);
     writeln;
   end;
 end.