http://poj.org/problem?id=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).
思路：(⊙v⊙)嗯和例题同理啊，从@开始，搜索到所有可以走到的地方，把那里改为一个值（@或者真值什么的），最后遍历一遍地图就好了。
1 #include <stdio.h>
2 char ch[24][24];
3 void solve(int H,int W);
4 void find1(int H,int W);
5 void dfs(int x,int y,int H,int W);
6 int main()
7 {
8 int H,W,i;
9 while(scanf("%d%d",&W,&H)!=EOF&&H!=0&&W!=0)
10 {
11 getchar();
12 for(i=0;i<H;i++)
13 gets(ch[i]);
14 find1(H,W);
15 solve(H,W);
16 }
17 return 0;
18 }
19
20 void solve(int H,int W)
21 {
22 int a=0;
23 for(int i=0;i<H;i++)
24 for(int k=0;k<W;k++)
25 if(ch[i][k]==‘@‘)
26 a++;
27 printf("%d\n",a);
28 }
29
30 void find1(int H,int W)
31 {
32 for(int i=0;i<H;i++)
33 for(int k=0;k<W;k++)
34 if(ch[i][k]==‘@‘)
35 {
36 dfs(i,k,H,W);
37 return;
38 }
39 }
40
41 void dfs(int x,int y,int H,int W)
42 {
43 int x1,y1,nx,ny;
44 ch[x][y]=‘@‘;
45 for(x1=-1;x1<=1;x1++)
46 {
47 nx=x+x1;
48 if(nx>=0&&nx<H&&ch[nx][y]!=‘#‘&&ch[nx][y]!=‘@‘)
49 dfs(nx,y,H,W);
50 }
51 for(y1=-1;y1<=1;y1++)
52 {
53 ny=y+y1;
54 if(ny>=0&&ny<W&&ch[x][ny]!=‘#‘&&ch[x][ny]!=‘@‘)
55 dfs(x,ny,H,W);
56 }
57 return;
58 }