poj 1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 25228		Accepted: 13605

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

/*
题意：给你一个n*m的矩阵，字符@是起点字符.是路字符#是墙，问你
从起点开始最多可以走多长的路（即可以走过多少个点，注意：墙不可穿过） 
题解：找到起点的坐标，然后向它的四个方向搜索，找到所有的可以走的路
每走过一个点，就在总路数上+1（同样注意将走过的路用#覆盖，避免重复） 
*/

#include<stdio.h> //poj1979
#include<string.h>
#include<algorithm>
#define MAX 110
#define INF 0x3f3f3f
using namespace std;
int n,m,j,i,t,k;
int sum;
char map[MAX][MAX];
int vis[MAX][MAX];
//int move[4][2]={1,0,-1,0,0,-1,0,1};
//此题不能用这个辅助数组，因为这个辅助数组走一步之后起点的
//（x,y）的坐标发生了变化，而走过的路已经被覆盖，导致无法再走 
void dfs(int x,int y)
{
	if(0<=x&&x<m&&0<=y&&y<n&&map[x][y]!='#')
	{
		map[x][y]='#';
		sum++;
		dfs(x+1,y);
		dfs(x-1,y);
		dfs(x,y-1);
		dfs(x,y+1);
		//这样搜索，由于搜索一次后（x,y）并没有改变就不会有上边的问题 
//		for(i=0;i<4;i++)
//		    dfs(x+move[i][0],y+move[i][1]);
	}
	return ;
}
int main()
{
    int x,y;
	while(scanf("%d%d",&n,&m),n|m)
	{
		sum=0; 
		for(i=0;i<m;i++)
		    scanf("%s",map[i]);
		for(i=0;i<m;i++)
			for(j=0;j<n;j++)		    
		        if(map[i][j]=='@')//找到起点坐标并记录下来 
		        {
		        	x=i;
		        	y=j;
		        }
		dfs(x,y);
		printf("%d
",sum);
	}
	return 0;
}