DFS基础题

hdu 1241 油田  裸DFS

题意：@代表油田 8个方向上还有@就相连 相当于求图中连通子图的个数
Sample Input
1 1 // n m
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

Sample Output
0
1
2
2

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<algorithm>
 using namespace std;

 int d[][] = {{,},{,-},{,},{,-},{,},{-,},{-,-},{-,}};
 char map[][] ;

 int n , m ;

 void dfs(int x , int y)
 {
     int fx , fy ;
     int i ;
     map[x][y] = '*' ;
     for (i =  ; i <  ; i++)
     {
         fx = x + d[i][] ;
         fy = y + d[i][] ;
         if (fx>= && fx<n && fy>= && fy<m && map[fx][fy] == '@')
         {
             dfs(fx,fy) ;
         }
     }
 }

 int main()
 {
     //freopen("in.txt","r",stdin) ;

    while (scanf("%d %d" , &n , &m) !=EOF)
    {
        if (n ==  && m == )
           break ;
        int i , j ;

        for (i =  ; i < n ; i++)
        {
            for (j =  ; j < m ; j++)
           {
               cin>>map[i][j];         //用scanf一直WA=.=
           }

        }
        int sum =  ;
        for (i =  ; i < n ; i++)
         for (j =  ; j < m ; j++)
        {
            if (map[i][j] == '@')
            {
                sum++ ;

                dfs(i,j) ;
            }
        }
        printf("%d\n" , sum) ;

    }

     return ;
 }

hdu 2952 

上下左右 4个方向相连 求连通子图的个数
Sample Input
2
4 4
#.#. //#能走
.#.#
#.##
.#.#
3 5
###.#
..#..
#.###

Sample Output
6
3

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<algorithm>
 using namespace std;

 int d[][] = {{,},{,-},{,},{-,}};
 char map[][] ;

 int n , m ;

 void dfs(int x , int y)
 {
     int fx , fy ;
     int i ;
     map[x][y] = '.' ;
     for (i =  ; i <  ; i++)
     {
         fx = x + d[i][] ;
         fy = y + d[i][] ;
         if (fx>= && fx<n && fy>= && fy<m && map[fx][fy] == '#')
         {
             dfs(fx,fy) ;
         }
     }
 }

 int main()
 {
    // freopen("in.txt","r",stdin) ;
    int T ;
    scanf("%d" , &T) ;

    while (T--)
    {
        scanf("%d %d" , &n , &m) ;
        if (n ==  && m == )
           break ;
        int i , j ;

        for (i =  ; i < n ; i++)
        {
            for (j =  ; j < m ; j++)
           {
               cin>>map[i][j];
           }

        }
        int sum =  ;
        for (i =  ; i < n ; i++)
         for (j =  ; j < m ; j++)
        {
            if (map[i][j] == '#')
            {
                sum++ ;
                dfs(i,j) ;
            }
        }
        printf("%d\n" , sum) ;

    }

     return ;
 }

hdu 1312 能走多少格  裸DFS

 

Sample Input
6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.

11 9
.#......... //.能走 #不能走 @起点
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output
45
59
6
13

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<algorithm>
 using namespace std;

 int d[][] = {{,},{,-},{,},{-,} } ;
 char map[][] ;

 int n , m ;
 int sum =  ;

 void dfs(int x , int y)
 {
     int fx , fy ;
     int i ;
     sum++ ;
     map[x][y] = '#';
     for (i =  ; i <  ; i++)
     {
         fx = x + d[i][] ;
         fy = y + d[i][] ;
         if (fx>= && fx<n && fy>= && fy<m && map[fx][fy] == '.')
         {
             dfs(fx,fy) ;
         }
     }
 }

 int main()
 {
    // freopen("in.txt","r",stdin) ;

    while (scanf("%d %d" , &m , &n) !=EOF)
    {
        if (n ==  && m == )
           break ;
        int i , j ;
        int bx , by ;
        sum =  ;

        for (i =  ; i < n ; i++)
        {
            for (j =  ; j < m ; j++)
           {
               cin>>map[i][j];
           }

        }
        for (i =  ; i < n ; i++)
        {
            for (j =  ; j < m ; j++)
           {
               if (map[i][j] == '@')
               {
                   bx = i ;
                   by = j ;
               }
           }

        }

        dfs(bx,by) ;

        printf("%d\n" , sum) ;

    }

     return ;
 }