HDU 1312 Red and Black(DFS,板子题,详解,零基础教你代码实现DFS)

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19731    Accepted Submission(s): 11991

Problem 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)

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

Source

Asia 2004, Ehime (Japan), Japan Domestic

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1312

分析：看到ZERO已经写了DFS，弱鸡也得补上此题，来一发！感谢梅大大的支持,代码的模版来自梅大大,我只是对主要步骤做了注释而已！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int _x[4]={1,0,-1,0};
 4 int _y[4]={0,-1,0,1};//相当于从第一象限开始顺时针转一周，对应的坐标轴上的点分别为(1,0),(0,-1),(-1,0),(0,1),就是数学意义上的方向向量
 5 char str[30][30];
 6 bool flag[30][30];//去判断点是否被标记过,DFS搜索一遍,被标记过记改点为1!
 7 int n,m,ans;
 8 void DFS(int x,int y)
 9 {
10     for(int ii=0;ii<4;ii++)
11     {
12         int i=x+_x[ii];
13         int j=y+_y[ii];
14         if(i<n&&j<m&&i>=0&&j>=0&&flag[i][j]==0&&str[i][j]=='.')//边界条件
15         {
16             ans++;
17             flag[i][j]=1;
18             DFS(i,j);
19         }
20     }
21 }
22 int main()
23 {
24    while(scanf("%d%d",&m,&n)!=EOF)
25    {
26        if(m==0&&n==0)
27         break;
28        int x,y;
29        memset(flag,0,sizeof(flag));
30        for(int i=0;i<n;i++)
31         scanf("%s",str[i]);
32        for(int i=0;i<n;i++)
33        {
34            for(int j=0;j<m;j++)
35            {
36                if(str[i][j]=='@')//找到起始点
37                {
38                    x=i;
39                    y=j;
40                }
41            }
42        }
43        ans=0;
44        flag[x][y]=1;
45        DFS(x,y);
46        printf("%d\n",ans+1);//第一个位置也算一个，所以+1
47    }
48    return 0;
49 }