杭电1312--Red and Black（Dfs）

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13090 Accepted Submission(s): 8116

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9

....#. 

.....# 

...... 

...... 

...... 

...... 

...... 

#@...# 

.#..#. 

11 9 

.#......... 

.#.#######. 

.#.#.....#. 

.#.#.###.#.

.#.#..@#.#. 

.#.#####.#. 

.#.......#. 

.#########. 

........... 

11 6 

..#..#..#.. 

..#..#..#.. 

..#..#..### 

..#..#..#@. 

..#..#..#.. 

..#..#..#.. 

7 7

..#.#.. 

..#.#.. 

###.### 

...@... 

###.### 

..#.#.. 

..#.#.. 

0 0

Sample Output

45 59 6 13

Source

Asia 2004, Ehime (Japan), Japan Domestic

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//比较基础的搜索题，但是做了很久。→ → ！！

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #define M 30
 5 using namespace std;
 6 
 7 int ans;
 8 int x, y;
 9 char map[M][M];
10 int dx[4]={0, 0, 1, -1}; 
11 int dy[4]={1, -1, 0, 0};
12 int a, b;
13 void dfs(int x, int y)
14 {
15     int m, n;
16     ans++;
17     map[x][y] = '@';
18     //printf("%d %d


", x, y);
19     for(int i = 0; i < 4; i++) //  
20     {
21         m = x + dx[i];
22         n = y + dy[i];
23         //printf("%d %d
", m, n); 
24         if(m >= 0 && m < b && n >= 0 && n < a && map[m][n] == '.')   //限制边界。 
25         {
26         //    printf("1");
27         //    printf(" %d %d
", m, n);
28             dfs(m, n);
29         }
30     }
31 }
32 
33 int main()
34 {
35     int i, j;
36     while(~scanf("%d %d", &a, &b), a|b)    
37     {
38         ans = 0;
39         for(i=0; i<b; i++)
40         {
41             for(j=0; j<a; j++)
42             {
43                 cin >> map[i][j];
44                 if(map[i][j] == '@')
45                 {
46                     x = i;
47                     y = j;
48                 }
49             }
50         }
51         //printf("%d %d

", x, y);
52         dfs(x, y);
53         printf("%d
", ans);
54     }
55     return 0;
56 }

注意：

 1 void dfs(int x, int y)
 2 {
 3     int nx, ny;
 4     ant++;
 5     map[x][y] = '@';
 6     for(int i = 0; i < 4; i++)
 7     {
 8         nx = x + acx[i];
 9         ny = y + acy[i];
10         if(nx >= 0 && nx < m && ny >= 0 && ny < n && map[nx][ny] == '.')
11         {
12             dfs(nx, ny);    
13         }  
14     }
15 }

important