3
3
ABA
CBC
3
BAA
CCB
3
BBB
BAB

Case #1: 3
Case #2: 3
Case #3: 2

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

struct NODE
{
    int val;
    int step;
};

bool vis[1 << 16];  //标记各位置是否达到目标色
char s[20];
int n;

int BFS()
{
    queue<NODE> q;
    NODE st, cur, next;
    memset(vis, false, sizeof(vis));
    vis[0] = true;
    st.val = 0;
    st.step = 0;
    q.push(st);
    while(!q.empty())
    {
        cur = q.front();
        q.pop();
        //各位都为1则涂色完成
        if(cur.val == (1 << 2 * n) - 1) 
            return cur.step;
        for(int i = 0; i < 2 * n; i++)
        {
            //如果当前位置已经涂色，则不作为拓展原点
            if((1 << i) & cur.val)
                continue;
            next = cur;
            next.step ++;
            int tmp = 0;
            //单行向右扩展
            int nn = i < n ? n : 2 * n;
            for(int j = i; j < nn; j++)
            {
                //若该点已经涂色则退出循环
                if((1 << j) & cur.val)
                    break;
                //若该点未涂色且颜色和目标色相同则上色
                if(s[j] == s[i])
                    tmp = tmp | (1 << j);
            }
            //单行向左扩展
            nn = i < n ? 0 : n;
            for(int j = i - 1; j >= nn; j--)
            {
                if((1 << j) & cur.val)
                    break;
                if(s[j] == s[i])
                    tmp = tmp | (1 << j);
            }
            //将所有状态分离，若出现新的状态则加入队列
            for(int j = tmp; j; j = tmp & (j - 1))
            {
                int val = cur.val | j;
                if(!vis[val])
                {
                    vis[val] = true;
                    next.val = val;
                    q.push(next);
                }
            }
            //双行扩展通过第一行来得到第二行，若当前对应
            //的第二行的位置已经涂色，则该列不作为拓展原点
            if(i >= n || (cur.val & (1 << i + n)))
                continue;
            tmp = 0;
            for(int j = i; j < n; j++)
            {
                if(((1 << j) & cur.val) || ((1 << (j + n)) & cur.val))
                    break;
                if(s[j] == s[i])
                    tmp |= (1 << j);
                if(s[j + n] == s[i])
                    tmp |= (1 << (j + n));
            }
            for(int j = i - 1; j >= 0; j--)
            {
                if(((1 << j) & cur.val) || ((1 << (j + n)) & cur.val))
                    break;
                if(s[j] == s[i])
                    tmp |= (1 << j);
                if(s[j + n] == s[i])
                    tmp |= (1 << (j + n));
            }
            for(int j = tmp; j; j = tmp & (j - 1))
            {
                int val = cur.val | j;
                if(!vis[val])
                {   
                    vis[val] = true;
                    next.val = val;
                    q.push(next);
                }
            }
        }
    }
    return 0;
}

int main()
{
    int T;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++)
    {
        scanf("%d", &n);
        scanf("%s %s", s, s + n); //将两行读做一行
        printf("Case #%d: %d\n", ca, BFS());
    }
}