bfs红与黑

Problem M

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 27   Accepted Submission(s) : 10

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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

思路：还是很简单的bfs。

代码实现：

#include 
#include 
struct queuenode{								//建结构
	int a, b;
	}queue[100000];

int count, sx, sy, tx, ty, head = 0,tail = 0;			//定义
int a, b, n, temp;
int dir[4][2] = { {1,0} , {0,1} , {-1,0} , {0,-1} };
bool map[110][110];
int m, k, i, j;
int inarea(int x,int y)					//判断是否在区域内
{
	return x >= 0 && y >= 0 && x < a && y < b;
}

void bfs(  )									//bfs
{
	tail = head;
	queue[head].a = sx;							//初始化队列头部
	queue[head].b = sy;
	map[sx][sy] = 1;
	while(head <= tail){	    				//在队列中有数据的情况下进行循环
		for(k = 0;k < 4; k++){				//四个方向
			tx = queue[head].a + dir[k][0];
			ty = queue[head].b + dir[k][1];
			if(inarea(tx, ty) && map[tx][ty]== 0){
				tail++;
				count++;
				queue[tail].a = tx;
				queue[tail].b = ty;
				map[tx][ty] = 1;
			}
		}
		head++;
	}
	return;
}
		
int main(void)									//主函数
{
	while(scanf("%d%d",&b,&a)==2&&b!=0){
		memset(map, 1, sizeof(map));
		while(scanf(" ")||scanf("\n")) ;
		for(i = 0;i < a; i++){					//扫描地图
			for(j = 0;j < b; j++){
				scanf("%c",&temp);
				if (temp == '#') map[i][j] = 1;
				else if (temp == '.') map[i][j] = 0;
				else if (temp == '@') {
					map[i][j] = 0;
					sx = i;
					sy = j;
				}
				else {
					j--;
					continue;
				}
			}
		}
		count = 1;
		bfs();									//进入bfs
		printf("%d\n",count);
	}
	return 0;
}