题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1312
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15186 Accepted Submission(s): 9401
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.
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
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6
13
59
6
13
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std;
#define N 25
char mp[N][N];
int go[][] = {{,},{-,},{,-},{,}}; int n,m;
bool ck(int x, int y){
if(x<n&&x>=&&y<m&&y>=) return true;
else return false;
}
bool vis[N][N];
void dfs(int x, int y, int &sum)
{
vis[x][y] = ;
bool fl = ;
for(int i = ; i < ; i++){
int xx = x + go[i][];
int yy = y + go[i][];
if(ck(xx,yy)&&!vis[xx][yy]&&mp[xx][yy]=='.'){
fl = ;
sum = sum+;
dfs(xx,yy,sum);
}
}
if(fl==) return;
}
int main()
{
while(~scanf("%d%d",&m,&n))
{
int x, y;
if(n==&&m==) break;
getchar();
for(int i = ; i < n; i++){
for(int j = ; j < m; j++){
scanf("%c",&mp[i][j]);
if(mp[i][j]=='@'){ x = i; y = j; }
}
getchar();
}
memset(vis,,sizeof(vis));
int sum = ;
dfs(x,y,sum);
printf("%d\n",sum+);
}
return ;
}