hdoj 1312 Red and Black【bfs】

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13953    Accepted Submission(s): 8644
dfs做法，参见http://blog.csdn.net/nailnehc/article/details/48007321

题目大意；给一个n*m的矩阵，‘@’为你当前站的位置，在矩阵中，‘ . ’表示可行走，‘ # ’表示不可行走，问从当前位置出发，能走到的格子有多少个【只能上下左右移动】；

已Accept代码【c++提交】

#include <cstdio>
#include <queue>
using namespace std;
char map[22][22];
int m, n;
int r, c;

int dx[] = {0, 0, 1, -1};
int dy[] = {-1, 1, 0, 0};
typedef struct node {
	int x, y;
}node;

queue <node> Q;
void BFS() {
	while(!Q.empty())
		Q.pop();
	node cur;
	cur.x = r;
	cur.y = c;
	Q.push(cur);
	int total = 1;
	map[cur.x][cur.y] = '#';
	while(!Q.empty()) {
		node tmp;
		tmp = Q.front();
		Q.pop();
		for(int i = 0; i < 4; i++) {
			int x = tmp.x + dx[i];
			int y = tmp.y + dy[i];
			if(x < 1 || x > n || y < 1 || y > m || map[x][y] == '#')
				continue;
			else {
				cur.x = x;
				cur.y = y;
				Q.push(cur);
				map[x][y] = '#';
				total++;
			}
		}
	}
	printf("%d\n", total);
}

int main() {
	while(scanf("%d%d", &m, &n), n | m) {
		for(int i = 1; i <= n; i++) {
			getchar();
			for(int j = 1; j <= m; j++) {
				scanf("%c", &map[i][j]);
				if(map[i][j] == '@') {
					r = i;
					c = j;
				}
			}
		}
		BFS();
	}
	return 0;
}