poj_1979

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\n", num);
    }
}