Hduoj1312【广搜】

/*Red and BlackTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10126    Accepted Submission(s): 6316Problem DescriptionThere 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. InputThe 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) OutputFor 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 Input6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0 Sample Output4559613 SourceAsia 2004, Ehime (Japan), Japan Domestic */ #include<stdio.h>#include<string.h>int vis[22][22], q[450], n, m, dx[4] = {-1, 1, 0 , 0}, dy[4] = {0, 0, -1, 1};char map[22][22];int bfs(int x, int y){memset(vis, 0, sizeof(vis));//初始化 int front  = 0, rear  = 0, num = 1, u;u = x*m+y; vis[x][y] = 1;//标记 q[rear++] = u;//入队起点 while(front < rear){u = q[front++];//逐个取出队列元素 x = u/m;y = u%m; for(int d = 0; d < 4; d++)//遍历四个方向 {int nx = x+dx[d], ny = y+dy[d];if(nx >= 0 && nx < n && ny >= 0 && ny < m && map[nx][ny] == '.' && !vis[nx][ny])//若边界内 且为黑 为访问 {num++;   //结果加1 vis[nx][ny] = 1;//标记以访问 int v = nx*m+ny; q[rear++] = v;//入队}}}return num;}int main(){int x, y;while(scanf("%d%d", &m, &n) != EOF && (n || m)){for(int i = 0 ; i < n; i++){scanf("%s", map[i]);if(strchr(map[i], '@')){for(int j = 0; j < m; j++)if(map[i][j] == '@')//保存起点 {x = i;y = j;break;}}}printf("%d\n", bfs(x, y));}return 0;}

题意：给出一个矩阵，包含起点和‘.', '#'， 求从起点开始，能够经过多少个‘.’，‘#’是墙。

思路：遍历矩阵，从起点开始遍历，遇到一个点就将答案加1，知道无法遍历或遍历完成为止，输出答案。