poj 1979Red and Black(BFS DFS)

 一道简单的搜索题，和poj2386差不多，DFS和BFS都可以,当然并查集也可以……

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 14518		Accepted: 7537

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

BFS；

#include<iostream>
#include<cstring>
using namespace std;
struct Node
{
	int x;
	int y;
}queue[7000];

char map[21][21];
int visited[21][21];
int w, h;
int ax[4]={-1, 1, 0, 0};
int ay[4]={0, 0, -1, 1};

int BFS(int x, int y)
{
	int num=1;//初始位置为黑色，所以初始值为1
	int front=0, rear=1;
	int i;
	int X,Y;
	visited[y][x]=1;//标记
	queue[front].x=x;
	queue[front].y=y;

	while( front != rear)
	{
		for(i=0; i<4; i++)
		{
			X=queue[front].x+ax[i];
			Y=queue[front].y+ay[i];
			if( !visited[Y][X] && (X>=0&&X<w) && (Y>=0&&Y<h))//未越界且未访问过，入队列
			{
				queue[rear].x=X;
				queue[rear].y=Y;
				visited[Y][X]=1;
				rear++;
				num++;
			}
		}
		front++;
	}
	return num;
}
int main()
{
	int i, j;
	int x, y;
	while( cin>>w>>h && w)
	{
		memset(visited, 0, sizeof(visited));
		for(i=0; i<h; i++)
			for(j=0; j<w; j++)
			{
				cin>>map[i][j];
				if( map[i][j]=='#')
					visited[i][j]=1;
				else if( map[i][j]=='@')//记录人的初始位置
				{
					x=j;y=i;
				}
			}

        cout<<BFS(x, y)<<endl;
	}
}

DFS:

#include<iostream>
#include<cstring>
using namespace std;

char map[21][21];
int visited[21][21];
int w, h;

int num;//初始位置为黑色，所以初始值为1
int ax[4]={-1, 1, 0, 0};
int ay[4]={0, 0, -1, 1};

void DFS(int x, int y)
{
	int i;
	if( visited[y][x] == 1 || (x<0||x>=w) || (y<0||y>=h)) //访问过，或越界，递归结束
        return ;
	else//未访问过,继续
    {
        visited[y][x]=1;//标记
        num++;
        for(i=0; i<4; i++)
		{
            DFS(x+ax[i], y+ay[i]);
		}
    }
}
int main()
{
	int i, j;
	int x, y;
	while( cin>>w>>h && w)
	{
	    num=0;//初始化初始值
		memset(visited, 0, sizeof(visited));
		for(i=0; i<h; i++)
			for(j=0; j<w; j++)
			{
				cin>>map[i][j];
				if( map[i][j]=='#')
					visited[i][j]=1;
				else if( map[i][j]=='@')//记录人的初始位置
				{
					x=j;y=i;
				}
			}
        DFS(x, y);
        cout<<num<<endl;
	}
}