Red and Black（BFS/DFS）

Red and Black

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

这道题用简单bfs/dfs都可以

#include<iostream>
#include<algorithm>
#include<queue>
#include<bits/stdc++.h>
using namespace std;

int m,n;
char a[205][205];
int minx;
bool vst[205][205];
int x,y;


int d[4][2]={0,1,0,-1,1,0,-1,0};
int ans=0;

void dfs(int x,int y)
{
	
	a[x][y]='#';
	ans++;
		
	for(int i=0;i<4;i++){
		int xx=x +d[i][0];
		int yy=y +d[i][1];
		if(xx>=0&&xx<m&&yy>=0&&yy<n&&vst[xx][yy]&&a[xx][yy]!='#'){
			
			//cout<<xx<<" "<<yy<<endl;
			dfs(xx,yy);
			
		}
	}
	
}

int main(){
	
	while(cin>>n>>m&&n&&m){
		ans=0;
		int xx,yy;
		memset(vst,1,sizeof(vst));
		for(int i=0;i<m;i++){
			for(int j=0;j<n;j++){
			cin>>a[i][j];
			if(a[i][j]=='@'){
				xx=i;
				yy=j;
			}
			}
		}
		//cout<<xx<<" &&& "<<yy<<endl;
		dfs(xx,yy);
		cout<<ans<<endl;
	}
	return 0;
}