Red and Black

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


Sample Output

45
59
6
13


Source

Asia 2004, Ehime (Japan), Japan Domestic
#include<iostream>
#include<algorithm>
using namespace std;
int fang[4][2]={1,0,0,1,-1,0,0,-1};
bool mapused[25][25];
char map[25][25];
int n,m;//列 行；
int number;
int dfs(int a,int b)
{
if(a>=0&&a<=m&&b>=0&&b<=n)
{

//判断有没有越界
if(map[a][b]=='#')//遇墙 返回
{
return 0;
}
else
if(mapused[a][b]==1)//遇到走过的 返回
{
return 0;
}
else//没走过 且不是墙；
{
mapused[a][b]=1;
number++;
int k;
for(k=0;k<4;k++)//搜索4个方向
{
dfs(a+fang[k][0],b+fang[k][1]);
}

}

}//越界结束
else return 0;

}
int main()
{

while(scanf("%d %d",&n,&m)!=EOF)//列 行
{
if(m==0&&n==0) break;
memset(mapused,0,sizeof(mapused));
number=0;
memset(map,'#',sizeof(map));//初始化
int i,j;
int starx,stary;
for(i=0;i<m;i++)
{

for(j=0;j<n;j++)
{
cin>>map[i][j];
if(map[i][j]=='@')
{
starx=i;//行
stary=j;//列

}

}

}//输入地图；
dfs(starx,stary);
printf("%d\n",number);

}

}

posted @ 2013-08-17 10:31  一只蚊子  阅读(155)  评论(0编辑  收藏  举报