一个美妙的栅栏

例题POJ   1037

代码：

#include<bits/stdc++.h>
using namespace std;
const int UP =0; const int DOWN =1;
const int MAXN = 25; 
long long C[MAXN][MAXN][2]; //C[i][k][DOWN] 是S(i)中以第k短的木棒打头的DOWN方案数,
							//C[i][k][UP] 是S(i)中以第k短的木棒打头的UP方案数,第k短指i根中第k短
void Init(int n) {//抽象化一下 
	memset(C,0,sizeof(C));
	C[1][1][UP] = C[1][1][DOWN] = 1; 
	for( int i = 2 ;i <= n; ++ i ) 
		for( int k = 1; k <= i; ++ k ) { //枚举第一根木棒的长度
			for( int M = k; M <i ; ++M ) //枚举第二根木棒的长度
				C[i][k][UP] += C[i-1][M][DOWN];
			for( int N = 1; N <= k-1; ++N ) //枚举第二根木棒的长度
				C[i][k][DOWN] += C[i-1][N][UP];	
	}
//总方案数是 Sum{ C[n][k][DOWN] + C[n][k][UP] } k = 1.. n;
}

void Print(int n, long long cc)
{
	long long skipped = 0; //已经跳过的方案数
	int seq[MAXN]; //最终要输出的答案
	int used[MAXN]; //木棒是否用过
	memset(used,0,sizeof(used));
	for( int i = 1; i<= n; ++ i ) { //依次确定每一个位置i的木棒序号
		long long oldVal = skipped;
		int k;
		int No = 0; //k是剩下的木棒里的第No短的,No从1开始算
		for( k = 1; k <= n; ++k ) { //枚举位置i的木棒 ，其长度为k 
			oldVal = skipped;
			if( !used[k]) {
				++ No; //k是剩下的木棒里的第No短的
				if( i == 1 ) 
					skipped += C[n][No][UP] + C[n][No][DOWN];
				else {
					if( k > seq[i-1] && ( i <=2 || seq[i-2]>seq[i-1])) 
					//合法放置
						skipped += C[n-i+1][No][DOWN];
					else if( k < seq[i-1] && (i<=2 || seq[i-2]<seq[i-1])) //合法放置
						skipped += C[n-i+1][No][UP];
				}
				if( skipped >= cc ) 
				break;	
			}
		}
		used[k] = true;
		seq[i] = k;
		skipped = oldVal;	
	}
	for( int i = 1;i <= n; ++i ) 
		if( i < n) printf("%d ",seq[i]);
		else printf("%d",seq[i]);
		printf("\n");
}
int main() {
	int T,n; 
	long long c;
	Init(20);//初始化20根棍子
	scanf("%d",&T); 
	while(T--) {
		scanf("%d %lld",&n,&c);
		Print(n,c);
	} 
	return 0;
}