/*** 并查集(简单):* 这题麻烦在各种方向的设定,其实用到的并查集很少,就是最后算集合数的* 方向设定好后,直接扫一遍n*m的图,对当前位置的处理就是 对可延伸方向(也就是水管能到下个方块的方向)* 的相邻方块判断,如果该延伸方向的水管同样能延伸过来,就把这两个方块的根节点合并起来。* 其实也就是并查集最基础的应用了。* 等扫完一遍图后,再去算最后的集合数就行了。* p.s.因为开的fa[]数组是一维数组,就有个从开始二维的map转换到一维的fa[](记录根节点的数组)数组的过程* 其实这一想也很简单, 就是 一维位置pos = i * m + j;*/#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <vector>#include <algorithm>#define DEBUG 0#define INF 0x7fffffff#define MAXS 2505#define LL long longusing namespace std;int n, m;int fa[MAXS], vis[MAXS],/** 顺序: 上、左、下、右. */ kinds[11][4] = {{1, 1, 0, 0}, {1, 0, 0, 1}, {0, 1, 1, 0}, {0, 0, 1, 1}, {1, 0, 1, 0}, {0, 1, 0, 1}, {1, 1, 0, 1}, {1, 1, 1, 0}, {0, 1, 1, 1}, {1, 0, 1, 1}, {1, 1, 1, 1} }, dir[4][2] = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};char map[51][51];int get_fa(int x) { if(x != fa[x]) fa[x] = get_fa(fa[x]); return fa[x];}void init(int n, int m) { for(int i = 1; i <= n * m; i ++) { fa[i] = i; vis[i] = 0; }}bool out(int x, int y) { if(x >= n || y >= m || x < 0 || y < 0) return true; return false;}void Union(int x, int y) { int xfa = get_fa(x), yfa = get_fa(y); fa[yfa] = xfa;}void solve(int n, int m) { for(int i = 0; i < n; i ++) { for(int j = 0; j < m; j ++) { int pos = m * i + j + 1; for(int k = 0; k < 4; k ++) { int nextx = i + dir[k][0], nexty = j + dir[k][1], nextPos = pos + m * dir[k][0] + dir[k][1]; if(!out(nextx, nexty)) { if(kinds[map[i][j] - 'A'][k] && kinds[map[nextx][nexty] - 'A'][(k + 2) % 4]) { if(get_fa(pos) != get_fa(nextPos)) { Union(pos, nextPos); } } } } } }}int main(){ while(scanf("%d%d", &n, &m)) { if(n == -1 || m == -1) break; init(n, m); for(int i = 0; i < n; i ++) scanf("%s", map[i]); solve(n, m); int ans = 0; for(int i = 1; i <= n * m; i ++) { if(!vis[get_fa(i)]) { ans ++; vis[get_fa(i)] = 1; } } printf("%d\n", ans); } return 0;}