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[0..20010] of node;
    p,cx,cy:array[0..20010] of longint;
    v:array[0..20010] 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<>0 do
    begin
      y:=e[i].po;
      if not v[y] then
      begin
        v[y]:=true;
        if (cy[y]=0) or (dfs(cy[y])=1) then
        begin
          cx[x]:=y;
          cy[y]:=x;
          exit(1);
        end;
      end;
      i:=e[i].next;
    end;
    exit(0);
  end;

begin
  readln(n);
  for i:=1 to n do
  begin
    read(d);
    x:=i+d;
    if x>n then x:=x-n;
    y:=i-d;
    if y<1 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:=0;
  for i:=n downto 1 do
    if cx[i]=0 then
    begin
      fillchar(v,sizeof(v),false);
      s:=s+dfs(i);
    end;
  if s<n then writeln('No Answer')
  else begin
    for i:=1 to n do
    begin
      write(cx[i]-1);
      if i<>n then write(' ');
    end;
  end;
end.