Red and Black

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

Total Submission(s): 10858    Accepted Submission(s): 6797

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0
```

Sample Output

```45
59
6
13
```

BFS和DFS

```#include <stdio.h>
#include <queue>
#include <string.h>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
char map[33][33];
int vis[33][33];
int n,m;
int sum;
struct node
{
int x,y;
}
f[333];
#define inf 0xfffffff
void bfs(int x,int y)
{
int i;
queue<node>q;
node st,ed;
st.x=x;
st.y=y;
q.push(st);
while(!q.empty())
{
st=q.front();
q.pop();
for(i=0;i<4;i++)
{
ed.x=st.x+dir[i][0];
ed.y=st.y+dir[i][1];
if(ed.x>=n ||ed.y>=m ||ed.x<0 ||ed.y<0 ||map[ed.x][ed.y]=='#' ||map[ed.x][ed.y]=='@')
continue;
map[ed.x][ed.y]='@';
sum++;
q.push(ed);
}
}
}
int main()
{
int i,j,x,y;
while(scanf("%d%d",&m,&n)!=EOF&&n!=0 &&m!=0)
{
for(i=0;i<n;i++)
scanf("%s",map[i]);
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if(map[i][j]=='@')
{
x=i;
y=j;
}
}
sum=1;
bfs(x,y);
printf("%d\n",sum);
}
return 0;
}
//广搜

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
char map[33][33];
int vis[33][33];
int n,m;
int sum;
#define inf 0xfffffff
void dfs(int x,int y)
{
int sx,sy,i;
sum++;
for(i=0;i<4;i++)
{
sx=x+dir[i][0];
sy=y+dir[i][1];
if(sx>=n ||sy>=m ||sx<0 ||sy<0 ||map[sx][sy]=='#' ||map[sx][sy]=='@')
continue;
map[sx][sy]='@';
dfs(sx,sy);
}
}
int main()
{
int i,j,x,y;
while(scanf("%d%d",&m,&n)!=EOF&&n!=0 &&m!=0)
{
for(i=0;i<n;i++)
scanf("%s",map[i]);
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if(map[i][j]=='@')
{
x=i;
y=j;
}
}
sum=0;
dfs(x,y);
printf("%d\n",sum);
}
return 0;
}
//深搜
//深搜和广搜都是只能讲满足的点放入继续搜索，所以搜过的点要把它变成不满足的点，满足的点可以继续再搜。这就是本题的关键```

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