题意：首先给你空闲的位置，可以跳过几个来吃掉几个，求最短的吃完所有的，且最后一个回到开始指定的位置

思路：BFS+HASH判重，对于每个位置有六个方向，当然有的是不能走的，加上map的判重就可以了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
using namespace std;
const int MAXN = 16;

int next[15][6] = {
	{-1, -1, -1, -1, 2, 3}, {-1, 1, -1, 3, 4, 5}, {1, -1, 2, -1, 5, 6}, 
	{-1, 2, -1, 5, 7, 8}, {2, 3, 4, 6, 8, 9}, {3, -1, 5, -1, 9, 10}, 
	{-1, 4, -1, 8, 11, 12}, {4, 5, 7, 9, 12, 13}, {5, 6, 8, 10, 13, 14}, 
	{6, -1, 9, -1, 14, 15}, {-1, 7, -1, 12, -1, -1}, {7, 8, 11, 13, -1, -1}, 
	{8, 9, 12, 14, -1, -1}, {9, 10, 13, 15, -1, -1}, {10, -1, 14, -1, -1, -1}
};

struct Statue {
	int st;
	int num;
	int way[2][MAXN];
	int cnt;
	Statue(int _st = (1<<16) - 1, int _num = 15) {
		st = _st;
		num = _num;
		cnt = 0;
	}
};
queue<Statue> q;
set<int> s;
int n;

int bfs() {
	s.clear();
	while (!q.empty()) 
		q.pop();
	q.push(Statue(((1<<16)-1)^(1<<(n-1)), 14));

	while (!q.empty()) {
		Statue tmp = q.front();
		q.pop();
		for (int i = 0; i < 15; i++) {		
			if ((tmp.st>>i)&1) {		
				for (int j = 0; j < 6; j++) {			
					if (next[i][j] != -1 && ((tmp.st>>(next[i][j]-1))&1)) {	
						int tt = i;
						Statue cur = tmp;
						while (tt >= 0 && (cur.st>>tt)&1) {
							cur.st -= (1<<tt);
							cur.num--;
							tt = next[tt][j] - 1;
						}
						cur.num++;
						if (tt < 0) 
							continue;
						cur.st |= (1<<tt);
						if (!s.count(cur.st)) {
							s.insert(cur.st);
							cur.way[0][cur.cnt] = i + 1;
							cur.way[1][cur.cnt] = tt + 1;
							cur.cnt++;
							if (cur.num == 1 && (cur.st>>(n-1))&1) {
								printf("%d\n%d %d", cur.cnt, cur.way[0][0], cur.way[1][0]);
								for (int i = 1; i < cur.cnt; i++) {
									printf(" %d %d", cur.way[0][i], cur.way[1][i]);
								}
								puts("");
								return 0;
							}
							q.push(cur);
						}
					}
				}
			}
		}
	}
	printf("IMPOSSIBLE\n");
	return 0;
}


int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		bfs();
	}
	return 0;
}