思路分析:

    n節火車, 對應2n次進出棧操作, 組織解空間樹的結構, 設解空間樹的每個葉結點對應一個進出棧方案, 按照字典序遍歷解空間樹的前20個葉結點即可, AC程式碼如下:

//CH1101_火車進棧
#include <cstdio>
#include <iostream>
#include <vector>
using namespace std;
const int MAX = 20;
int t;
vector<int> res;//res[0..n - 1]:出棧方案
vector<int> in, out;//in[0...]當前棧中火車棧底到棧頂, out[0...]當前未進棧火車
//, out[0]編號最小, 且初始時刻out[0...n - 1] = 1...n 
void solve(){
	if(t == MAX) return;
	if(in.empty() && out.empty()){
		for(int i = 0; i < res.size(); ++i) cout << res[i]; 
		cout << endl, ++t; return;
	}
	if(!in.empty()){
		int x = in.back();
		res.push_back(x), in.pop_back(), solve(), in.push_back(x), res.pop_back();
	}
	if(out.empty()) return; 
	int x = out.back();
	in.push_back(x), out.pop_back(), solve(), in.pop_back(), out.push_back(x);
}
int main(){
	int n; scanf("%d", &n); for(int i = n; i >= 1; --i) out.push_back(i);
	solve();
	return 0;
}