zoj 3861 Valid Pattern Lock（dfs）

Valid Pattern Lock

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

给出数字的位置 在上面连线来解锁 

给出所要相连的数字  连接的位置要是相连的 如果中间隔了数字 那么这个数字必须是之前已经经过的

dfs 找出所有符合情况的组合

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 500010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char ch;
    int a = 0;
    while((ch = getchar()) == ' ' | ch == '\n');
    a += ch - '0';
    while((ch = getchar()) != ' ' && ch != '\n')
    {
        a *= 10;
        a += ch - '0';
    }
    return a;
}

void Print(int a)    //Êä³öÍâ¹Ò
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}
int n;
int mp[15][15];
int vis[15],ex[15];//分别表示 是否已经经过这个数字 这个数字是否给出
int num;
int ans[MAXN][15];
int res[15];

void init()
{
    MEM(mp,0);
    mp[1][3]=mp[3][1]=2;
    mp[4][6]=mp[6][4]=5;
    mp[7][9]=mp[9][7]=8;
    mp[1][7]=mp[7][1]=4;
    mp[2][8]=mp[8][2]=5;
    mp[3][9]=mp[9][3]=6;
    mp[1][9]=mp[9][1]=5;
    mp[3][7]=mp[7][3]=5;
}

void dfs(int cnt)
{
    if(cnt>n)
    {
        for(int i=1;i<=n;i++)
            ans[num][i]=res[i];
        num++;
//        return;
    }
    else
    {
        for(int i=1;i<=9;i++)
        {
            if((vis[i]==1)&&(ex[i]==1)&&(vis[mp[i][res[cnt-1]]]==0)&&(ex[mp[i][res[cnt-1]]]==1))
            {
                vis[i]=0;
                res[cnt]=i;
                dfs(cnt+1);
                vis[i]=1;
                res[cnt]=0;
            }
        }
    }
}

int main()
{
//    fread;
    init();
    int tc;
    scanf("%d",&tc);
    while(tc--)
    {
        scanf("%d",&n);
        MEM(vis,0); MEM(ex,0);
        ex[0]=1;
        for(int i=0;i<n;i++)
        {
            int x;
            scanf("%d",&x);
            vis[x]=ex[x]=1;
        }
        num=0;
//        res[0]=0;
        dfs(1);
        printf("%d\n",num);
        for(int i=0;i<num;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(j>1) printf(" ");
                printf("%d",ans[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}