题目：
| Red and Black| Time Limit: 1000MS | | Memory Limit: 30000K |
| ---------------------------- | - | ------------------------ |
| Total Submissions: 32833 | | Accepted: 17852 |DescriptionThere 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. InputThe 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. OutputFor 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 Input6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0Sample Output```
45 59 6 13
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思路：
题意很简单,"#"代表墙，"."可以走，"@"代表出发点，问最多能走多少步（不走重复）,简单深搜，直接放代码
代码：
```cpp
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int w,h,x1,y1,maxx;
char map[50][50];
int vis[50][50];
int go[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
void dfs(int x,int y)
{
for(int i=0; i<4; i++)
{
int xx=x+go[i][0];
int yy=y+go[i][1];
if(xx>=0&&xx<h&&yy>=0&&yy<w&&vis[xx][yy]==0&&map[xx][yy]=='.')
{
vis[xx][yy]=1;
dfs(xx,yy);
maxx++;
}
}
}
int main()
{
while(~scanf("%d%d",&w,&h)&&(w||h))
{
maxx=1;
mem(vis,0);
for(int i=0; i<h; i++)
{
scanf("%s",map[i]);
for(int j=0; j<w; j++)
if(map[i][j]=='@')
{
x1=i;
y1=j;
}
}
dfs(x1,y1);
printf("%d\n",maxx);
}
return 0;
}
\
