题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1312
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
思路：玩家从‘@’开始出发，遇到’#‘走不了，遇到’.’能继续走，问能经过整张图的多少个点（包括该玩家的起始的位置）。代码如下：
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int n,m,step,x1,y1; // step用于记录步数
char map[30][30];
int vis[30][30]; //标记数组
int dir[4][2]={1,0,-1,0,0,1,0,-1}; //方向数组
void DFS(int x,int y)
{
if(x<0||x>=n||y<0||y>=m||map[x][y]=='#'||vis[x][y]) //设置递归出口
return;
else
{
step++; //一旦执行else语句，所走步数就会+1
for(int i=0;i<4;i++)
{
vis[x][y]=1;
x1=x+dir[i][0];
y1=y+dir[i][1];
DFS(x1,y1);
}
}
}
int main()
{
while(cin>>m>>n&&n&&m)
{
step=0;
memset(vis,0,sizeof(vis));
memset(map,0,sizeof(map));
for(int i=0;i<n;i++)
cin>>map[i];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(map[i][j]=='@')
{
DFS(i,j);
//break;
}
}
}
cout<<step<<endl;
}
return 0;
}ps:你喜欢大海，而哦喜欢你！