题目：
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
分析：这个题目是一个bfs问题；只要在输入的时候确定了@的坐标，并从这个坐标开始四个方向广度搜索，并同时做好已访问标记，也就是记忆性搜索（剪枝）；
代码：
#include<stdio.h> //#include<stdio.h> int num, row, col; int direction[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; //四个方向 char tiles[25][25]; int v[25][25]; void bfs(int dx, int dy) { int i, x, y; // if(!v[dx][dy]) { // return; //剪枝 // } for(i = 0; i < 4; ++i) { x = dx + direction[i][0]; y = dy + direction[i][1]; if(!v[x][y] && x >= 0 && y >= 0 && x < row && y < col && tiles[x][y] == '.') { num++; v[x][y] = 1; bfs(x, y); } } return; } int main() { int dx, dy, i, j; while(scanf("%d%d", &col, &row)) { getchar(); if(col == 0 && row == 0) break; for(i = 0; i < row; ++i) { for(j = 0; j < col; ++j) { v[i][j] = 0; tiles[i][j] = getchar(); if(tiles[i][j] == '@') { dx = i; dy = j; } } getchar(); } num = 0; bfs(dx, dy); ++num; printf("%d ", num); } }