1 #include<cstdio>//hdu1198 dfs
 2 #include<cstring>
 3
 4 int well[11][5] = { {1,1,1,0,0},{1,0,1,1,0},{0,1,1,0,1},{0,0,1,1,1},{1,0,1,0,1},{0,1,1,1,0},{1,1,1,1,0},{1,1,1,0,1},{0,1,1,1,1},{1,0,1,1,1},{1,1,1,1,1}};
 5
 6 const int ca[5][2] = {0,1,1,0,1,1,1,2,2,1}, mov[4][2] = {1,0,-1,0,0,1,0,-1};
 7 int m,n,sum, map[550][550],id[550][550];
 8
 9 void build(int set, int x, int y)
10 {
11     for(int k = 0; k<5; k++)
12     {
13         if(well[set][k])
14         {
15             map[3*x+ca[k][0]][3*y+ca[k][1]] = 1;
16         }
17     }
18 }
19
20 void dfs(int x, int y, int iden)
21 {
22     id[x][y] = iden;
23     for(int i = 0; i<4; i++)
24     {
25         int xx = x + mov[i][0], yy = y + mov[i][1];
26         if(xx<0 || xx>3*m-1 || yy<0 || yy>3*n-1 ) continue;
27         if(map[xx][yy] && !id[xx][yy])
28         dfs(xx,yy,iden);
29     }
30 }
31
32 int main()
33 {
34     char init[5005];
35     while(scanf("%d%d",&m,&n) && (m!=-1 || n!=-1 ))
36     {
37         memset(map,0,sizeof(map));
38         memset(id,0,sizeof(id));
39         for(int i = 0; i<m; i++)
40         {
41             scanf("%s",init);
42             for(int j = 0; j<n; j++)
43             build(init[j]-65,i,j);
44         }
45         sum = 0;
46         for(int i = 0; i<=3*m-1; i++)
47         {
48
49             for(int j = 0; j<=3*n-1; j++)
50             {
51                 //printf("%d ",map[i][j]);
52                 if(map[i][j] && !id[i][j])
53                 {
54                     dfs(i,j,++sum);
55                 }
56             }
57             //putchar('\n');
58         }
59
60         printf("%d\n",sum);
61
62     }
63     return 0;
64 }

 1 #include<cstdio>//hdu1198 并查集 每个格子的标号为i*n+j，初始father也为i*n+j，以次来作为合并集合的依据
 2 #include<cstring>
 3
 4 int n,m,father[55][55];
 5 char map[55][55];
 6 char type[11][5]={"1010","1001","0110","0101","1100","0011",
 7                "1011","1110","0111","1101","1111"};
 8
 9 int find(int x)
10 {
11     return (x==father[x/n][x%n])?x:find(father[x/n][x%n]);
12 }
13
14 int unit(int a, int b)
15 {
16     a = find(a);
17     b = find(b);
18     if(a!=b)//合并集合
19         father[a/n][a%n] = b;
20 }
21
22
23 int judge(int i, int j)
24 {   //判断是否与左边相连
25     if(j>0 && type[map[i][j]-'A'][2]=='1' && type[map[i][j-1]-'A'][3]=='1')
26         unit(i*n+j,i*n+j-1);
27     //判断是否与上边相连
28     if(i>0 && type[map[i][j]-'A'][0]=='1' && type[map[i-1][j]-'A'][1]=='1')
29         unit(i*n+j,(i-1)*n+j);
30 }
31
32 int main()
33 {
34     while(scanf("%d%d",&m,&n) && (m!=-1 || n!=-1))
35     {
36         for(int i = 0; i<m; i++)
37         {
38             scanf("%s",&map[i]);
39             for(int j = 0; j<n; j++)
40             {   //初始化为自己
41                 father[i][j] = i*n+j;
42             }
43         }
44
45         //并查集过程
46         for(int i = 0; i<m; i++)
47         for(int j = 0; j<n; j++)
48         judge(i,j);
49
50         int sum = 0;
51         for(int i = 0; i<m; i++)
52         for(int j = 0; j<n; j++)//看看有多少个“根节点”
53             if(father[i][j]==i*n+j) sum++;
54
55         printf("%d\n",sum);
56     }
57
58 }