给一个m*n的矩阵，数字是0或是1，统计其中八连块的个数，和《算法竞赛入门经典》6.4.1黑白图像一样，应该是那个的原型吧。
代码如下：
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1 #include <cstdio> 2 #include <cstring> 3 4 const int maxn = 105; 5 int mat[maxn][maxn], vis[maxn][maxn]; 6 7 void dfs(int x, int y) 8 { 9 if(vis[x][y] || !mat[x][y]) return; 10 vis[x][y] = 1; 11 dfs(x-1, y-1); dfs(x-1, y); dfs(x-1, y+1); 12 dfs(x, y-1); dfs(x, y+1); 13 dfs(x+1, y-1); dfs(x+1, y); dfs(x+1, y+1); 14 } 15 16 int main() 17 { 18 #ifdef LOCAL 19 freopen("in", "r", stdin); 20 #endif 21 int m, n; 22 while(scanf("%d%d", &m , &n) != EOF && m) 23 { 24 char s[maxn]; 25 memset(mat, 0, sizeof(mat)); 26 memset(vis, 0, sizeof(vis)); 27 for(int i = 0; i < m; i++) 28 { 29 scanf("%s", s); 30 for(int j = 0; j < n; j++) 31 { 32 if(s[j] == '*') mat[i+1][j+1] = 0; 33 else if(s[j] == '@') mat[i+1][j+1] = 1; 34 } 35 } 36 int cnt = 0; 37 for(int i = 1; i <= m; i++) 38 for(int j = 1; j <= n; j++) 39 if(!vis[i][j] && mat[i][j]) 40 { 41 cnt++; 42 dfs(i, j); 43 } 44 printf("%d\n", cnt); 45 } 46 return 0; 47 }
上面是用递归的方式实现的，也可用显式栈来代替递归，代码如下：
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1 #include <cstdio> 2 #include <cstring> 3 #include <stack> 4 using namespace std; 5 6 const int maxn = 105; 7 int mat[maxn][maxn], vis[maxn][maxn]; 8 int m, n; 9 10 const int dir[][2] = { {1, 1}, {1, 0}, {1, -1}, {0, 1}, {0, -1}, {-1, 1}, {-1, 0}, {-1,-1} }; 11 12 struct Pos 13 { 14 int x, y; 15 }; 16 17 void dfs(int x, int y) 18 { 19 stack<Pos> s; 20 Pos t; 21 t.x = x; 22 t.y = y; 23 s.push(t); 24 while(!s.empty()) 25 { 26 t = s.top(); 27 s.pop(); 28 int x = t.x, y = t.y; 29 for(int i = 0; i < 8; i++) 30 { 31 int nx, ny; 32 nx = x + dir[i][0]; 33 ny = y + dir[i][1]; 34 if(nx >= 0 && nx < m && ny >= 0 && ny < n) 35 if(mat[nx][ny] && !vis[nx][ny]) 36 { 37 t.x = nx; 38 t.y = ny; 39 s.push(t); 40 vis[nx][ny] = 1; 41 } 42 } 43 } 44 } 45 46 47 int main() 48 { 49 #ifdef LOCAL 50 freopen("in", "r", stdin); 51 #endif 52 while(scanf("%d%d", &m, &n) != EOF && m) 53 { 54 char s[maxn]; 55 memset(mat, 0, sizeof(mat)); 56 memset(vis, 0, sizeof(vis)); 57 for(int i = 0; i < m; i++) 58 { 59 scanf("%s", s); 60 for(int j = 0; j < n; j++) 61 if(s[j] == '@') mat[i][j] = 1; 62 } 63 int cnt = 0; 64 for(int i = 0; i < m; i++) 65 for(int j = 0; j < n; j++) 66 if(!vis[i][j] && mat[i][j]) 67 { 68 cnt++; 69 dfs(i, j); 70 } 71 printf("%d\n", cnt); 72 } 73 return 0; 74 }