搜索技术——广度优先搜索（BFS）

广度优先搜索和深度优先搜索是基本的暴力技术，常用于解决图，树的遍历问题。
算法思路为：在每个节点的可扩展处都扩散一遍，如无法继续扩展便停下，等到所有的分列都停止之后广度优先搜索便结束。
广度优先搜索的扩散一般都是用队列来进行实现。

代码实现(…为自定义操作)：

void BFS(....)
{
	queue<type> q;
	..............
	q.push(n);
	while(!q.empty())
	{
		q.pop();
		............
		if(....)
		{
			.......
			q.push(m);
		}
	}
}

（如果有误还请指正）

具体实践

Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit:
65536/32768 K (Java/Others) Total Submission(s): 35748 Accepted
Submission(s): 21693

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

Recommend Eddy

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

int width, high;
int num;//记录总数
char room[21][21];

struct node
{
	int x;
	int y;
};

int dir[4][2] = { {-1,0},{0,-1},{1,0},{0,1} };//往四面扩展

void BFS(int dx, int dy)
{
	queue<node> q;
	node start, next;
	num = 1;
	start.x = dx;
	start.y = dy;
	q.push(start);
	while (!q.empty())
	{
		start = q.front();
		q.pop();
		for (int i = 0; i < 4; i++)
		{
			next.x = start.x + dir[i][0];
			next.y = start.y + dir[i][1];
			if (next.x >= 0 && next.x < width && next.y >= 0 && next.y < high && room[next.x][next.y] == '.')
			{
				room[next.x][next.y] = '#';
				q.push(next);
				num++;
			}
		}
	}
}

int main()
{
	int x, y;
	while (cin >> width >> high)
	{
		for (int i = 0; i < high; i++)
		{
			for (int j = 0; j < width; j++)
			{
				cin >> room[j][i];
				if (room[j][i] == '@')//设置初始点
				{
					x = j;
					y = i;
				}
			}
		}
		num = 0;
		BFS(x, y);
		cout << num << endl;
	}
	return 0;
}

（题目略水）