hdu-1312-Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 13464 Accepted Submission(s): 8338



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
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).

[align=left]Sample Input[/align]

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

[align=left]Sample Output[/align]

45
59
6
13

[align=left]Source[/align]
Asia 2004, Ehime (Japan), Japan
Domestic

[align=left]Recommend[/align]
Eddy | We have carefully selected several similar problems for you: 1241 1010 1372 1242 1253

题目分析：

最简单的dfs 就是让你 从@出发最对能走多少步 可以四个方向走但是遇到#不能走

#include<cstdio>
#include<cstring>
char a[22][22];
int x,y;
int n,m;
int cnt;
void dfs(int x,int y)
{
if(a[x][y]=='#') return;
if(x>=1&&x<=m&&y>=1&&y<=n)
{
a[x][y]='#';// 把走过的路都标记成墙不能再走了
cnt++;//  每走一步就加一
dfs(x+1,y);// 向下遍历
dfs(x-1,y);//  向上遍历
dfs(x,y-1);//  向左遍历
dfs(x,y+1);// 向右遍历    就是每个方向都走一遍直到遇到 # 转方向
}
}
int main()
{

while(~scanf("%d%d",&n,&m)&&n|m)
{
int i,j;
memset(a,0,sizeof(a));
for(i=1;i<=m;i++)
{
scanf("%s",a[i]+1);// 记住这里一定要加一因为 你的初始位置是  1  1  如果不加仪的话初始位置就         //是1  0了   因为这里选择的是一行一行的输入每行是一个字符串  字符串的首地址就是  0 所以这里要加1  变成 1 1
for(j=1;j<=n;j++)
{
if(a[i][j]=='@')
{
x=i;
y=j;
}
}
}
cnt=0;
dfs(x,y);
printf("%d\n",cnt);
}
return 0;
}