《挑战》2.1 POJ POJ 1979 Red and Black （简单的DFS）

B - Red and Black

Submit Status Practice POJ 1979

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)         The end of the input is indicated by a line consisting of two zeros.

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

题目大意：

　　就是说，对于一个h*w的网格，每次从‘@’开始，求最多能走多少个‘.’，每次只能够向四个方向进行扩展。

解题思路：

　　直接dfs，然后用book数组随时记录目前这个点有没有走过，如果这个点是'.'并且没有走过的话，那么我们就res++，由于是多组数据，每次输出结束后，记着memset,

复杂度：

　　O(4*w*h)

代码：

 1 # include<cstdio>
 2 # include<iostream>
 3 # include<cstring>
 4 
 5 using namespace std;
 6 
 7 # define MAX 23
 8 
 9 char grid[MAX][MAX];
10 int book[MAX][MAX];
11 int nxt[4][2] = {{1,0},{0,-1},{-1,0},{0,1} };
12 int res;
13 int w,h;
14 
15 int can_move ( int x,int y )
16 {
17     if ( x>=0&&x<h&&y>=0&&y<w )
18     {
19         if ( grid[x][y]=='.'&&book[x][y]==0 )
20             return 1;
21     }
22     return 0;
23 }
24 
25 void dfs ( int x,int y )
26 {
27     book[x][y] = 1;
28     for ( int i = 0;i < 4;i++ )
29     {
30         int n_x = x+nxt[i][0], n_y = y+nxt[i][1];
31         if ( can_move ( n_x,n_y ) )
32         {
33             res++;
34             book[n_x][n_y] = 1;
35             dfs(n_x,n_y);
36         }
37     }
38 }
39 
40 int main(void)
41 {
42     while ( scanf("%d%d",&w,&h)!=EOF )
43     {
44         res = 1;
45         if ( w==0&&h==0 )
46             break;
47         for ( int i = 0;i < h;i++ )
48             scanf("%s",grid[i]);
49         for ( int i = 0;i < h;i++ )
50         {
51             for ( int j = 0;j < w;j++ )
52             {
53                 if ( grid[i][j]=='@' )
54                 {
55                     book[i][j] = 1;
56                     dfs(i,j);
57                 }
58             }
59         }
60 
61         printf("%d
",res);
62         memset(book,0,sizeof(book));
63     }
64 
65 
66 
67     return 0;
68 }