ZOJ-3110

说白了就是一道连通块搜索，前天好像做过一个二维的，这题是三维的，没什么难度，我用的BFS，看大牛题解说是用DFS，没想到怎么做。最后那个输出格式有点坑，本来结果早就算出来了，还格式化了半天。。话说我怎么越看我的python代码越丑，怎么看都不习惯，还是C的代码比较清爽，可惜这题用到队列，懒的用C搞了

import sysdef get_adjs(src, i, j, l, m, p, k, code):    r = []    if i + 1 < m and src[i + 1][j][l] == code:        r.append([i + 1, j, l])    if i - 1 >= 0 and src[i - 1][j][l] == code:        r.append([i - 1, j, l])    if j + 1 < p and src[i][j + 1][l] == code:        r.append([i, j + 1, l])    if j - 1 >= 0 and src[i][j - 1][l] == code:        r.append([i, j - 1, l])    if l + 1 < k and src[i][j ][l + 1] == code:        r.append([i, j , l + 1])    if l - 1 >= 0 and src[i][j ][l - 1] == code:        r.append([i, j , l - 1])    return rdef bfs(src, i, j, l, m, p, k):    code = src[i][j][l]    src[i][j][l] = '*'    queue = [[i, j, l]]    total = 0    while(len(queue) != 0):        curr = queue[0]        del queue[0]        total += 1        for adj in get_adjs(src, curr[0], curr[1], curr[2], m, p, k, code):            src[adj[0]][adj[1]][adj[2]] = '*'            queue.append([adj[0], adj[1], adj[2]])    return total    stdin = sys.stdincount = 0while(True):    line = stdin.readline()    if not line:        break    count += 1    pkm = line.strip().split()    p = int(pkm[0])    k = int(pkm[1])    m = int(pkm[2])    src = []    for i in range(0, m):        temp = []        for j in range(0, p):            temp.append(list(stdin.readline().strip()))        src.append(temp)        stdin.readline()        res = {'a':[], 'b':[], 'c':[], 'd':[]}    for i in range(0, m):        for j in range(0, p):            for l in range(0, k):                if src[i][j][l] != '*':                    res[src[i][j][l]].append(bfs(src, i, j, l, m, p, k))    if count != 1:        print    print 'Case %d:' % (count)    abcd = ['a', 'b', 'c', 'd']    for cc in abcd:        s = cc        for i in sorted(res[cc], reverse=True):            s += ' ' + str(i)        print s