Gym 100952E&&2015 HIAST Collegiate Programming Contest E. Arrange Teams【DFS+剪枝】

E. Arrange Teams

time limit per test:2 seconds

memory limit per test:64 megabytes

input:standard input

output:standard output

Syrian Collegiate Programming Contest (SCPC) is the qualified round for the Arab Collegiate Programming Contest. Each year SCPC organizers face a problem that wastes a lot of time to solve it, it is about how should they arrange the teams in the contest hall.

Organizers know that they have t teams and n*m tables allocated in n rows and m columns in the contest hall. They don't want to put teams in a random way. Each year SCPC chief judge puts a list of paired teams (list of a,b teams) that should not sit next to each other (because they are so good or so bad!).

if pair (a,b) is in chief judge list, this means that:

- team number a should not sit in-front of team number b

- team number b should not sit in-front of team number a

- team number a should not sit right to team number b

- team number b should not sit right to team number a

Organizers wastes a lot of time to find a good team arrangement that satisfy all chief judge needs. This year they are asking you to write a program that can help them.

Input

First line contains number of test cases. The first line in each test case contains three numbers: (1  ≤  n,m  ≤  11) and (1  ≤  t  ≤  10). Second line in each test case contains (0  ≤  p  ≤  40) number of pairs. Then there are p lines, each one of them has two team numbers a and b (1  ≤  a,b  ≤  t) where a is different than b.

Output

For each test case, print one line contains the total number of teams arrangements that satisfy all chief judge needs (We guarantee that it will be less than 9,000,000 for each test case). If there is no suitable arrangements print "impossible".

Examples

Input

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

Output

2
impossible

Note

In test case 1 there are 2 teams and 3 tables in one row at the contest hall. There are only one pair (1,2), so there are 2 solutions:

team1 then empty table then team2

team2 then empty table then team1

In test case 2 there are 4 tables in 2 rows and 2 columns, and there are 4 teams. There is no arrangement that can satisfy chief judge needs.

题目链接：http://codeforces.com/gym/100952/problem/E

分析：dfs、剪枝

首先用dp[vis[a][b]][k]布尔数组双向的记录那些vis[i][j-1],vis[i][j+1],vis[i-1][j],vis[i+1][j];a=i-1,i+1,i,i；b=j,j,j-1,j+1；

然后inline void dfs(int k)表示当前正在处理队伍k,然后遍历所有i, j如果没有访问过,并且满足条件则dfs(k + 1)

直到顺利的把所以的t个队伍都填进去了,那一个分枝才ans++; return;

剪枝以后的复杂度不大算的出来,但看数据大小 (1  ≤  n,m  ≤  11) and (1  ≤  t  ≤  10) 这样做一般可以AC

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 inline int read()
 4 {
 5     int x=0,f=1;
 6     char ch=getchar();
 7     while(ch<'0'||ch>'9')
 8     {
 9         if(ch=='-')
10             f=-1;
11         ch=getchar();
12     }
13     while(ch>='0'&&ch<='9')
14     {
15         x=x*10+ch-'0';
16         ch=getchar();
17     }
18     return x*f;
19 }
20 inline void write(int x)
21 {
22     if(x<0)
23     {
24         putchar('-');
25         x=-x;
26     }
27     if(x>9)
28         write(x/10);
29     putchar(x%10+'0');
30 }
31 int n,m,t,ans;
32 int vis[15][15];
33 bool dp[15][15];
34 inline void DFS(int k)
35 {
36     if(k==t+1)
37     {
38         ans++;
39         return;
40     }
41     for(int i=1;i<=n;i++)
42     {
43         for(int j=1;j<=m;j++)
44         {
45             if(vis[i][j])
46                 continue;
47             if(!dp[vis[i-1][j]][k]&&!dp[vis[i+1][j]][k]&&!dp[vis[i][j-1]][k]&&!dp[vis[i][j+1]][k])
48             {
49                 vis[i][j]=k;
50                 DFS(k+1);
51                 vis[i][j]=0;
52             }
53         }
54     }
55 }
56 int main()
57 {
58     int T,q,x,y;
59     T=read();
60     while(T--)
61     {
62         ans=0;
63         memset(dp,0,sizeof(dp));
64         memset(vis,0,sizeof(vis));
65         n=read();
66         m=read();
67         t=read();
68         q=read();
69         for(int i=0;i<q;i++)
70         {
71             x=read();
72             y=read();
73             dp[x][y]=1;
74             dp[y][x]=1;
75         }
76         DFS(1);
77         if(ans!=0)
78             write(ans),printf("\n");
79         else printf("impossible\n");
80     }
81     return 0;
82 }