bzoj1562

很明显是二分图匹配，关键是怎么求字典序最小

想到两种做法，首先是直接匹配，然后从第一位贪心调整

第二种是从最后一个倒着匹配，每次匹配都尽量选小的，这样一定能保证字典序最小

 type node=record
        po,next:longint;
      end;

 var e:array[..] of node;
     p,cx,cy:array[..] of longint;
     v:array[..] of boolean;
     i,n,len,x,y,s,d:longint;

 procedure add(x,y:longint);
   begin
     inc(len);
     e[len].po:=y;
     e[len].next:=p[x];
     p[x]:=len;
   end;

 function dfs(x:longint):longint;
   var i,y:longint;
   begin
     i:=p[x];
     while i<> do
     begin
       y:=e[i].po;
       if not v[y] then
       begin
         v[y]:=true;
         if (cy[y]=) or (dfs(cy[y])=) then
         begin
           cx[x]:=y;
           cy[y]:=x;
           exit();
         end;
       end;
       i:=e[i].next;
     end;
     exit();
   end;

 begin
   readln(n);
   for i:= to n do
   begin
     read(d);
     x:=i+d;
     if x>n then x:=x-n;
     y:=i-d;
     if y< then y:=y+n;
     if x>y then
     begin
       add(i,x);
       add(i,y);
     end
     else begin
       add(i,y);
       if x<>y then add(i,x);
     end;
   end;
   s:=;
   for i:=n downto  do
     if cx[i]= then
     begin
       fillchar(v,sizeof(v),false);
       s:=s+dfs(i);
     end;
   if s<n then writeln('No Answer')
   else begin
     for i:= to n do
     begin
       write(cx[i]-);
       if i<>n then write(' ');
     end;
   end;
 end.