nyoj-27-水池数目

 1 //题目地址：http://acm.nyist.net/JudgeOnline/status.php?pid=27
 2 //解题思路：裸的dfs, 遍历每个格子dfs访问当前格子的相领格子
 3 #include <stdio.h>
 4 #include <string.h>
 5 
 6 int map[110][110], vis[110][110];
 7 int dx[4] = {0, 0, -1, 1};
 8 int dy[4] = {-1, 1, 0, 0};
 9 
10 void dfs(int x, int y){
11         int i;
12         if(map[x][y] == 0 || vis[x][y] == 1)    return ;        //访问过，或者不存在水池则返回
13         vis[x][y] = 1;
14         for(i = 0; i  <4; i++){
15                 dfs(x + dx[i], y + dy[i]);
16         }
17 //        dfs(x - 1, y);
18 //        dfs(x +1, y);
19 //        dfs(x, y - 1);
20 //        dfs(x, y + 1);
21 }
22 
23 int main(){
24         int t, n, m, i, j;
25         int ans;
26         scanf("%d", &t);
27         while(t--){
28                 scanf("%d %d", &n, &m);
29                 memset(map, 0, sizeof(map));
30                 memset(vis, 0, sizeof(vis));
31                 for(i = 1; i <= n; i++){
32                         for(j = 1; j <= m; j++){
33                                 scanf("%d", &map[i][j]);
34                         }
35                 }
36                 ans = 0;
37                 for(i = 1; i <= n; i++){
38                         for(j = 1; j <= m; j++){
39                                 if(map[i][j] == 1 && vis[i][j] == 0){
40                                         ans++;
41                                         dfs(i, j);
42                                 }
43                         }
44                 }
45                 printf("%d\n", ans);
46         }
47         return 0;
48 }