PKU1167 the buses

         这是我印象中敲的第一个dfs-id吧……弱弱地撒花……

         题目大家都清楚，我觉得在中国是个信息学的书上都有……注意搜索顺序！头脑中有一棵搜索树的话，就不难发现，如果搜索当前车属于原有线路还是新线路，随着树深的增加，枝条就越多，剪枝的效果就越好！

         迭代加深就是比普通的深度优先搜索多了几行代码而已，用于限制树的深度。而且得到的第一个可行解就是最优解！在状态空间比较庞大，或不便于记录时，可以用它代替广度优先搜索！

program the_buses;type rec=record mark,interval,first:longint;end;var data:array[0..300] of longint; sum:array[0..60] of longint; route:array[0..20] of rec; n,max,ans,dd:longint; flag:boolean;procedure prework;begin max:=0; dd:=1; flag:=false; fillchar(sum,sizeof(sum),0);end;procedure init;var i,w:longint;begin readln(n); for i:=1 to n do begin read(w); if w>max then max:=w; data[i]:=w; inc(sum[w]); end;end;procedure dfs(dep,tmp:longint);var i,w,time,tt:longint;begin if dep>n then begin for i:=1 to tmp do if route[i].mark=1 then exit; flag:=true; ans:=tmp; exit; end; w:=data[dep]; if sum[w]=0 then dfs(dep+1,tmp); for i:=1 to tmp do with route[i] do if mark=1 then begin interval:=w-first; if interval<=first then continue; mark:=2; time:=w; while (time<=max) and (sum[time]>0) do begin dec(sum[time]); inc(time,interval); end; if time>max then dfs(dep+1,tmp); if flag then exit; mark:=1; tt:=time-interval; time:=w; while time<=tt do begin inc(sum[time]); inc(time,interval); end; end; if (tmp>=dd) or (w shl 1>=max) then exit; inc(tmp); with route[tmp] do begin mark:=1; first:=w; end; dec(sum[w]); dfs(dep+1,tmp); if flag then exit; inc(sum[w]);end;procedure outit;begin writeln(ans);end;begin assign(input,'a.in'); reset(input); assign(output,'a.out'); rewrite(output); prework; init; while not flag do begin dfs(1,0); inc(dd); end; outit; close(input); close(output);end.