Problem 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.
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)
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
解题思路：简单深搜，水过！
AC代码：
#include<iostream>
using namespace std;
const int maxn=;
int col,row,cnt,si,sj,dir[][]={{-,},{,},{,-},{,}};char mp[maxn][maxn];
void dfs(int x,int y){
if(x<||y<||x>=row||y>=col||mp[x][y]=='#')return;
mp[x][y]='#';cnt++;//标记为'#'，并且计数器加1
for(int i=;i<;++i)
dfs(x+dir[i][],y+dir[i][]);//直接搜索四个方向
}
int main(){
while(cin>>col>>row&&(col+row)){
for(int i=;i<row;++i){
for(int j=;j<col;++j){
cin>>mp[i][j];
if(mp[i][j]=='@'){si=i;sj=j;}
}
}
cnt=;mp[si][sj]='.';dfs(si,sj);//从'@'开始搜索
cout<<cnt<<endl;
}
return ;
}