[POJ - 1979] Red and Black（DFS）

题目传送门：点击传送
题目来源：Japan 2004 Domestic
Time Limit: 1000MS
Memory Limit: 30000K

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.（输出能到达的’.'数量）

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 （不能走）
‘@’ - a man on a black tile(appears exactly once in a data set)
（起点）
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

解析：
题意：从’@‘开始走能到达多少个’.’，注意’#‘是障碍，被它围起来的’.'是走不了的。
（走：上下左右）

代码

#include <stdio.h>
char a[20][20];
int W,H,cnt=0;//H为行 W为列 cnt为可到达点('.')的数量
int main(void)
{
    int dfs(int x,int y);
    int i,j,m,n;
    while(~scanf("%d%d",&W,&H))
    {
        if(W==0||H==0) continue;//特殊情况
        cnt=0;//初始化操作
        getchar();
        for(i=0;i<H;i++)
        {
            for(j=0;j<W;j++)
            {
                scanf("%c",&a[i][j]);
                if(a[i][j]=='@')
                    m=i,n=j;//记录起点
            }
            getchar();//清除换行
        }
        dfs(m,n);//从起点开始走
        printf("%d\n",cnt);
    }
}

int dfs(int x,int y)
{
    if(x>=0&&x<H&&y>=0&&y<W&&(a[x][y]=='.'||a[x][y]=='@'))
    {   
        a[x][y]='#';//走过的不再走
        cnt++;
        dfs(x+1,y);
        dfs(x-1,y);
        dfs(x,y+1);
        dfs(x,y-1);
    }
    return 0;
}

碎碎念
[POJ - 2386] Lake Counting (简单的DFS)
两题的dfs都简单应用于图的搜索，而且做法比较相似，对走过的点进行覆盖
（有的题目需要新建一个标记数组来记录是不是走过）

DFS的话，理解思想和模板代码后的话会相对轻松