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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6
13
思路：就是从@开始上下左右4个方向把.全部找到，dfs就好
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5 char str[25][25];
6 int vis[25][25],w,h,sum;
7 int x[]={-1,1,0,0};
8 int y[]={0,0,-1,1};
9 void dfs(int i,int j)
10 {
11 for(int k=0; k<4; k++)
12 {
13 if(i+x[k]<h&&i+x[k]>=0&&j+y[k]<w&&j+y[k]>=0&&vis[i+x[k]][j+y[k]]==0&&str[i+x[k]][j+y[k]]!=‘#‘)
14 {
15 sum++;
16 vis[i+x[k]][j+y[k]]=1;
17 dfs(i+x[k],j+y[k]);
18 }
19 }
20 }
21 int main()
22 {
23 #ifdef CDZSC_OFFLINE
24 freopen("in.txt","r",stdin);
25 #endif
26 int i,j;
27 while(scanf("%d%d",&w,&h)&&w!=0&&h!=0)
28 {
29 memset(vis,0,sizeof(vis));
30 sum=1;
31 for(i=0; i<h; i++)
32 {
33 scanf("%s",str[i]);
34 }
35 for(i=0; i<h; i++)
36 {
37 for(j=0; j<w; j++)
38 {
39 if(str[i][j]==‘@‘)
40 {
41 vis[i][j]=1;
42 dfs(i,j);
43 }
44 }
45 }
46 printf("%d\n",sum);
47 }
48 return 0;
49 }