POJ-1979 Red and Black

Red and Black

POJ - 1979

时限: 1000MS

内存: 30000KB

64位IO格式: %I64d & %I64u

提交 状态

已开启划词翻译

问题描述

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. 

输入

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. 

输出

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).

样例输入

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

样例输出

45
59
6
13

 //#include <bits/stdc++.h>   ，POJ不支持这种头文件 
#include <iostream>
using namespace std;

const int N=25;
char str[N][N];
int w,h,cnt,tmpx,tmpy;

int dfs(int x,int y)
{
	if(x>=0&&x<h&&y>=0&&y<w&&(str[x][y]=='.'||str[x][y]=='@'))
	{
		str[x][y]='#';		//访问过后标志为'.'，这样后面的就不会在访问它了 
		cnt++;
		dfs(x+1,y);
		dfs(x,y-1);
		dfs(x-1,y);
		dfs(x,y+1);
	}
	return cnt;
}
int main()
{
	while(cin>>w>>h&&(w+h!=0))
	{
		cnt=0;
		for(int i=0;i<h;i++)
			for(int j=0;j<w;j++)
			{
				cin>>str[i][j];
				if(str[i][j]=='@')
				{
					tmpx=i;
					tmpy=j;
				}
			}
		cout<<dfs(tmpx,tmpy)<<endl;
					
	}
	return 0;
}