POJ1037 A decorative fence

题意

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A decorative fence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8477		Accepted: 3244

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 dierent 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

给定长度依次为1到n的木棒n个， 摆放规则为除了两边的木棒，剩下的木棒必须要比相邻的两个都高或者都低。求从小到大排列的第k个序列.

分析

参照superheroQAQ的题解。

其实在做过一些组合一类的题目之后这种题并不算太难 但是还是给了人很多启发并且有很多需要注意的地方.要想求第k个排列 f[i][j]表示由i根棍组成的j高度开头的种类数 这个是比较容易想到的 但是题目要求 波浪形！我记得noip考过一个简单的波浪形的题 于是再加一维状态 表示w型还是m形 求答案的时候就是dfs 这道题的dfs类似求排列 但有很多限制条件.

注意：

一根棍只能出现一次 并且我们要在意的是相对高度而不是实际高度 实际上就是说

dp[i][j][1/0]是可以由dp[i-1][j][0/1]转移过来的 举个例子：

dp[2][2][0] 2 1 ->dp[3][2][1] 2 3 1

时间复杂度\(O(n^2)\)

代码

#include<iostream>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;

co int N=21;
int n;
ll m,f[N][N][2];
void prework(){
	f[1][1][0]=f[1][1][1]=1;
	for(int i=2;i<N;++i)
		for(int j=1;j<=i;++j){
			for(int p=j;p<=i-1;++p) f[i][j][0]+=f[i-1][p][1];
			for(int p=1;p<=j-1;++p) f[i][j][1]+=f[i-1][p][0];
		}
}
bool used[N];
int main(){
	prework();
	for(int t=read<int>();t--;){
		read(n),read(m);
		memset(used,0,sizeof used);
		int last,k;
		for(int j=1;j<=n;++j){
			if(f[n][j][1]>=m) {last=j,k=1;break;}
			else m-=f[n][j][1];
			if(f[n][j][0]>=m) {last=j,k=0;break;}
			else m-=f[n][j][0];
		}
		used[last]=1;
		printf("%d",last);
		for(int i=2;i<=n;++i){
			k^=1;
			int j=0;
			for(int len=1;len<=n;++len){
				if(used[len]) continue;
				++j;
				if(k==0&&len<last||k==1&&len>last){
					if(f[n-i+1][j][k]>=m) {last=len;break;}
					else m-=f[n-i+1][j][k];
				}
			}
			used[last]=1;
			printf(" %d",last);
		}
		puts("");
	}
	return 0;
}