Pku1037 A decorative fence 動態規劃+遞推

A decorative fence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3468		Accepted: 1122

Description

Richard just finished building his new house. Now the only thing the house misses is a cute little wooden fence. He had no idea how to make a wooden fence, so he decided to order one. Somehow he got his hands on the ACME Fence Catalogue 2002, the ultimate resource on cute little wooden fences. After reading its preface he already knew, what makes a little wooden fence cute.
A wooden fence consists of N wooden planks, placed vertically in a row next to each other. A fence looks cute if and only if the following conditions are met:
�The planks have different lengths, namely 1, 2, . . . , N plank length units.
�Each plank with two neighbors is either larger than each of its neighbors or smaller than each of them. (Note that this makes the top of the fence alternately rise and fall.)
It follows, that we may uniquely describe each cute fence with N planks as a permutation a1, . . . , aN of the numbers 1, . . . ,N such that (any i; 1 < i < N) (ai − ai−1)*(ai − ai+1) > 0 and vice versa, each such permutation describes a cute fence.
It is obvious, that there are many di erent cute wooden fences made of N planks. To bring some order into their catalogue, the sales manager of ACME decided to order them in the following way: Fence A (represented by the permutation a1, . . . , aN) is in the catalogue before fence B (represented by b1, . . . , bN) if and only if there exists such i, that (any j < i) aj = bj and (ai < bi). (Also to decide, which of the two fences is earlier in the catalogue, take their corresponding permutations, find the first place on which they differ and compare the values on this place.) All the cute fences with N planks are numbered (starting from 1) in the order they appear in the catalogue. This number is called their catalogue number.

After carefully examining all the cute little wooden fences, Richard decided to order some of them. For each of them he noted the number of its planks and its catalogue number. Later, as he met his friends, he wanted to show them the fences he ordered, but he lost the catalogue somewhere. The only thing he has got are his notes. Please help him find out, how will his fences look like.

Input

The first line of the input file contains the number K (1 <= K <= 100) of input data sets. K lines follow, each of them describes one input data set.
Each of the following K lines contains two integers N and C (1 <= N <= 20), separated by a space. N is the number of planks in the fence, C is the catalogue number of the fence.
You may assume, that the total number of cute little wooden fences with 20 planks fits into a 64-bit signed integer variable (long long in C/C++, int64 in FreePascal). You may also assume that the input is correct, in particular that C is at least 1 and it doesn抰 exceed the number of cute fences with N planks.

Output

For each input data set output one line, describing the C-th fence with N planks in the catalogue. More precisely, if the fence is described by the permutation a1, . . . , aN, then the corresponding line of the output file should contain the numbers ai (in the correct order), separated by single spaces.

Sample Input

2
2 1
3 3

Sample Output

1 2
2 3 1

Source

CEOI 2002

題目大意：要你求滿足以下條件的，字典序爲k，長度爲n的數列：

這個數列由1~n組成，對於數列中的每個數Ai,滿足(Ai-Ai-1)*(Ai-Ai+1)>0。

分析：這個問題類似於康託展開，可以用類似的方法解決。方法是，用f[i]表示以i爲第一個數字的滿足條件的數列有多少個，然後就能一步一步推出整個數列。具體方法如下：

定義f[i,j]表示以長度爲i，以j爲第一個數字，第二個數比第一個數字大的數列有多少個。g[i,j]的定義類似，不同的是第二個數字比第一個數字小而已。

這兩個數組都可以用動態規劃求，方程也不難寫。

f[i,j]=∑g[i-1,k]（j<=k<=n-1）

g[i,j]=∑f[i-1,k]（1<=k<=j-1）

爲什麼呢？這個狀態並沒有考慮數列中出現了哪些數字，實際上也沒有必要考慮。因爲，對於一個數列，若第一個數字確定了，則後面的數字可以看作一個長度少1的全部數字加1的數列，因而得到了狀態轉移方程。

根據f和g數組，推導每位的數字的方法可以參看程序，應該不難理解。

注意的是，第一位要考慮比第二位大還是比第二位小，而後面每位只需根據上一次的決策來判斷要比上一次大還是小就行了。

 var f,g:array[1..20,1..20] of int64; a:array[1..20] of longint; b:array[1..20] of boolean; n,t,i,j:longint; k:int64; procedure init; var i,j,k:longint; begin f[1,1]:=1; g[1,1]:=1; for i:=2 to 20 do for j:=1 to i do begin for k:=j to i-1 do f[i,j]:=f[i,j]+g[i-1,k]; for k:=1 to j-1 do g[i,j]:=g[i,j]+f[i-1,k]; end; end; procedure dfs(x,l,r,col:longint); var i:longint; begin if x>n then exit else if col=0 then for i:=l to r do if g[n-x+1,i]>=k then begin a[x]:=i; dfs(x+1,1,i-1,1); exit; end else k:=k-g[n-x+1,i] else for i:=l to r do if f[n-x+1,i]>=k then begin a[x]:=i; dfs(x+1,i,n-x,0); exit; end else k:=k-f[n-x+1,i]; end; procedure print; var i,x:longint; function find(x:longint):longint; var temp,i:longint; begin temp:=0; for i:=1 to n do if not b[i] then begin inc(temp); if temp=x then exit(i); end; end; begin fillchar(b,sizeof(b),false); for i:=1 to n do begin x:=find(a[i]); b[x]:=true; write(x,' '); end; writeln; end; begin init; readln(t); for i:=1 to t do begin fillchar(a,sizeof(a),0); readln(n,k); for j:=1 to n do begin if g[n,j]>=k then begin a[1]:=j; dfs(2,1,j-1,1); break; end else k:=k-g[n,j]; if f[n,j]>=k then begin a[1]:=j; dfs(2,j,n-1,0); break; end else k:=k-f[n,j]; end; print; end; end.