HDU(1312)

Red and Black

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

Total Submission(s): 12211 Accepted Submission(s): 7601

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.

here 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

//上下左右搜索查看与起点能连通的个数

#include <stdio.h>
#include <iostream>
using namespace std;
char c[21][21];
int vis[21][21];
int n,m;
void dfs(int x,int y)
{
if(x<0||y<0||x>=m||y>=n) return;
if(vis[x][y]==1) return;
if(c[x][y]=='.'||c[x][y]=='@')  //起点也算一个
{
vis[x][y]=1;  //连通的标记为1
dfs(x+1,y);
dfs(x-1,y);
dfs(x,y+1);
dfs(x,y-1);
}
}
int main()
{
while(~scanf("%d%d",&n,&m)&&(n+m))
{
for(int i=0;i<21;i++)
{
for(int k=0;k<21;k++)
{
c[i][k]=0;
vis[i][k]=0;
}
}
int x,y;
for(int i=0;i<m;i++)
{
for(int k=0;k<n;k++)
{
cin>>c[i][k];
if(c[i][k]=='@')  //记录起点
{
x=i;
y=k;
}
}
}
dfs(x,y);
int cnt=0;
for(int i=0;i<m;i++)  //遍历一遍看看有几个标记的
for(int k=0;k<n;k++)
if(vis[i][k]==1)
cnt++;
printf("%d\n",cnt);
}
return 0;
}