《挑战》2.1 POJ POJ 1979 Red and Black （简单的DFS）

B - Red and Black

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1979

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

题目大意：

　　就是说，对于一个h*w的网格，每次从‘@’开始，求最多能走多少个‘.’，每次只能够向四个方向进行扩展。

解题思路：

　　直接dfs，然后用book数组随时记录目前这个点有没有走过，如果这个点是'.'并且没有走过的话，那么我们就res++，由于是多组数据，每次输出结束后，记着memset,

复杂度：

　　O(4*w*h)

代码： 

# include<cstdio>
# include<iostream>
# include<cstring>

using namespace std;

# define MAX 23

char grid[MAX][MAX];
int book[MAX][MAX];
int nxt[4][2] = {{1,0},{0,-1},{-1,0},{0,1} };
int res;
int w,h;

int can_move ( int x,int y )
{
    if ( x>=0&&x<h&&y>=0&&y<w )
    {
        if ( grid[x][y]=='.'&&book[x][y]==0 )
            return 1;
    }
    return 0;
}

void dfs ( int x,int y )
{
    book[x][y] = 1;
    for ( int i = 0;i < 4;i++ )
    {
        int n_x = x+nxt[i][0], n_y = y+nxt[i][1];
        if ( can_move ( n_x,n_y ) )
        {
            res++;
            book[n_x][n_y] = 1;
            dfs(n_x,n_y);
        }
    }
}

int main(void)
{
    while ( scanf("%d%d",&w,&h)!=EOF )
    {
        res = 1;
        if ( w==0&&h==0 )
            break;
        for ( int i = 0;i < h;i++ )
            scanf("%s",grid[i]);
        for ( int i = 0;i < h;i++ )
        {
            for ( int j = 0;j < w;j++ )
            {
                if ( grid[i][j]=='@' )
                {
                    book[i][j] = 1;
                    dfs(i,j);
                }
            }
        }

        printf("%d
",res);
        memset(book,0,sizeof(book));
    }



    return 0;
}