zoj 3861 Valid Pattern Lock

Valid Pattern Lock Time Limit: 2 Seconds      Memory Limit: 65536 KB

Pattern lock security is generally used in Android handsets instead of a password. The pattern lock can be set by joining points on a 3 × 3 matrix in a chosen order. The points of the matrix are registered in a numbered order starting with 1 in the upper left corner and ending with 9 in the bottom right corner.

A valid pattern has the following properties:

• A pattern can be represented using the sequence of points which it's touching for the first time (in the same order of drawing the pattern). And we call those points as active points.
• For every two consecutive points A and B in the pattern representation, if the line segment connecting A and B passes through some other points, these points must be in the sequence also and comes before A and B, otherwise the pattern will be invalid.
• In the pattern representation we don't mention the same point more than once, even if the pattern will touch this point again through another valid segment, and each segment in the pattern must be going from a point to another point which the pattern didn't touch before and it might go through some points which already appeared in the pattern.

Now you are given n active points, you need to find the number of valid pattern locks formed from those active points.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer n (3 ≤ n ≤ 9), indicating the number of active points. The second line contains n distinct integers a₁, a₂, … a_n (1 ≤ a_i ≤ 9) which denotes the identifier of the active points.

Output

For each test case, print a line containing an integer m, indicating the number of valid pattern lock.

In the next m lines, each contains n integers, indicating an valid pattern lock sequence. The m sequences should be listed in lexicographical order.

Sample Input

1
3
1 2 3

Sample Output

4
1 2 3
2 1 3
2 3 1
3 2 1

题意：手机图案解锁，给定n个数字，求可以得到的图案有多少种。其中有一些数字需要借助别的数字才能到达。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int vis[300],a[11],mp[11][11];
int ans[400000][11],f[11];
int n,len;

void init()
{
	mp[1][3] = 2; mp[3][1] = 2;
	mp[1][7] = 4; mp[7][1] = 4;
	mp[1][9] = 5; mp[9][1] = 5;
	mp[2][8] = 5; mp[8][2] = 5;
	mp[3][7] = 5; mp[7][3] = 5;
	mp[3][9] = 6; mp[9][3] = 6;
	mp[4][6] = 5; mp[6][4] = 5;
	mp[7][9] = 8; mp[9][7] = 8;
}
void print()
{
	for (int i = 0; i < n; i++)
	{
		ans[len][i] = f[i];
	}
	len++;
}
void dfs(int cur,int x)
{
	if (x == n)
	{
		print();
		return;
	}
	for (int i = 0; i <n; i++)
	{
		if (vis[a[i]] == 0 && (mp[a[i]][cur] == 0 || vis[mp[a[i]][cur]] == 1))//判断方式跟枚举一样
		{
			vis[a[i]] = 1;
			f[x] = a[i];
			dfs(a[i], x + 1);//a[i]作为cur传递，相当于两个两个比较
			vis[a[i]] = 0;//有多种，递归不成功改回来
		}
	}
}

int main()
{
	int t;
	scanf("%d", &t);
	memset(mp, 0, sizeof(mp));
	init();
	while (t--)
	{
		len = 0;
		memset(vis, 0, sizeof(vis));
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
			scanf("%d", &a[i]);
		sort(a, a + n);			//字典序排序
		dfs(0,0);			//第一次传递cur为0时相当于还没开始比较两个，必定成功
		printf("%d\n", len);
		for (int i = 0; i < len; i++)
		{
			for (int j = 0; j < n-1; j++)
				printf("%d ", ans[i][j]);
			printf("%d\n",ans[i][n-1]);//输出有点严格，不能多空格
		}
	}
	return 0;
}