HDU(1312)Red and Black(简单dfs)

Red and Black
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit Status

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

Sample Output

45
59
6
13

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
char map[100][100];
int sum,n,m;
void dfs(int x,int y)
{
if(map[x][y]=='#')
return;
if(x>=0&&x<m&&y>=0&&y<n)//保证在范围内搜索
{
sum++;
map[x][y]='#';
dfs(x,y-1);
dfs(x,y+1);
dfs(x-1,y);
dfs(x+1,y);
}

}
int main()
{
while(scanf("%d%d",&n,&m)&&n||m)
{int x,y;
memset(map,0,sizeof(map));
for(int i=0;i<m;i++)
{getchar();
for(int j=0;j<n;j++)
{
scanf("%c",&map[i][j]);
if(map[i][j]=='@')
{
x=i;
y=j;
}
}}
sum=0;
dfs(x,y);
printf("%d\n",sum);
}
}