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：

 #include <bits/stdc++.h>
 using namespace std;
 int _x[]={,,-,};
 int _y[]={,-,,};//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
 char str[][];
 bool flag[][];//去判断点是否被标记过,DFS搜索一遍,被标记过记改点为1!
 int n,m,ans;
 void DFS(int x,int y)
 {
     for(int ii=;ii<;ii++)
     {
         int i=x+_x[ii];
         int j=y+_y[ii];
         if(i<n&&j<m&&i>=&&j>=&&flag[i][j]==&&str[i][j]=='.')//边界条件
         {
             ans++;
             flag[i][j]=;
             DFS(i,j);
         }
     }
 }
 int main()
 {
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(m==&&n==)
         break;
        int x,y;
        memset(flag,,sizeof(flag));
        for(int i=;i<n;i++)
         scanf("%s",str[i]);
        for(int i=;i<n;i++)
        {
            for(int j=;j<m;j++)
            {
                if(str[i][j]=='@')//找到起始点
                {
                    x=i;
                    y=j;
                }
            }
        }
        ans=;
        flag[x][y]=;
        DFS(x,y);
        printf("%d\n",ans+);//第一个位置也算一个，所以+1
    }
    return ;
 }