POJ 1979 Red and Black

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

思路一：（DFS）

#include<iostream>
#include<cstring>
using namespace std;
const int N = 30;
int row, col, sx, sy;
int vis[N][N];
char a[N][N];
int dir[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
void dfs(int x, int y){
    for(int i = 0; i < 4; i++){
        int tx = dir[i][0] + x, ty = dir[i][1] + y;
        if(tx < 0 || tx > row - 1 || ty < 0 || ty > col - 1)    continue;
        if(!vis[tx][ty] && a[tx][ty] == '.'){
            vis[tx][ty] = 1;
            dfs(tx, ty);
        }
    }
}
int main(){
    while(cin >> col >> row){
        if(col == 0 && row == 0)    break;
        int cnt = 0;
        memset(vis, 0, sizeof(vis));
        for(int i = 0; i < row; i++){
            for(int j = 0; j < col; j++){
                cin >> a[i][j];
                if(a[i][j] == '@'){
                    sx = i;    sy = j;
                }
            }
        }
        vis[sx][sy] = 1;
        dfs(sx, sy);
        for(int i = 0; i < row; i++){
            for(int j = 0; j < col; j++){
                if(vis[i][j] == 1)    cnt++;
            }
        }
        cout << cnt << endl;    
    }
    return 0;
}

思路二：（BFS）

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
struct node{
    int x, y;
}que[2003];
char a[23][23];
int book[25][25], cnt, h, w;
int pos[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
void bfs(int rx, int ry){
    int head = 1, tail = 1;
        que[tail].x = rx;    que[tail].y = ry;
        book[rx][ry] = 1;
        tail++;    cnt++;
        while(head < tail){
            for(int i = 0; i < 4; i++){
                int tx = que[head].x + pos[i][0];
                int ty = que[head].y + pos[i][1];
                if(tx < 0 || tx > h - 1 || ty < 0 || ty > w - 1)    continue;
                if(a[tx][ty] == '.' && book[tx][ty] == 0){
                    cnt++;    
                    book[tx][ty] = 1;
                    que[tail].x = tx;    que[tail].y = ty;
                    tail++;
                }
            }
            head++;
        }
}
int main(){
    while(cin >> w >> h){
        if(w == 0 && h == 0)    break;
        memset(a, 0, sizeof(a));
        memset(book, 0, sizeof(book));
        cnt = 0;
        for(int i = 0; i < h; i++){
            cin >> a[i];
        }
        int rx, ry;
        for(int i = 0; i < h; i++){
            for(int j = 0; j < w; j++){
                if(a[i][j] == '@'){
                    rx = i;
                    ry = j;
                }
            }
        }
        bfs(rx, ry);
        cout << cnt << endl;
    }
    return 0;
}