HDU 1312 Red and Black （dfs）

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

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17773    Accepted Submission(s):
10826

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

题目大意： “.”代表黑地板，“#”代表红地板，“@”代表人的位置，人可以经过黑地板不能经过红地板，问在给定的图中人一共可以经过多少块地板

解题思路：深搜，在每个位置都搜索该位置的上下左右四个位置，将图中能够经过的点全部搜一遍

AC代码：

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <stack>
 #include <queue>
 using namespace std;
 int w,h;
 int a[][] = {{-,},{,},{,-},{,}};  //位置数组
 char str[][];
 int f[][];  //标记该位置是否走过
 int sum;
 void dfs(int x,int y)
 {
     f[x][y] = ;
     for (int i = ; i < ; i ++)
     {
         int x1=x + a[i][];
         int y1=y + a[i][];
         if (x1 >=  && x1 < h && y1 >=  && y1 < w && str[x1][y1]!='#' && f[x1][y1] == )
         {    //判断边界、是否是不能经过的红地板、该地板是否已经经过
             sum ++;
             dfs(x1,y1);
         }
     }
 }
 int main ()
 {
     int i,j,x,y;
     while (scanf("%d%d",&w,&h),w&&h)
     {
         for (i = ; i < h; i ++)
             scanf("%s",str[i]);

         memset(f,,sizeof(f));
         for (i = ; i < h; i ++)
         for (j = ; j < w; j ++)
             if (str[i][j] == '@') //找出人的位置
             {
                 x = i;
                 y = j;
                 break;
             }
         sum = ;
         dfs(x,y);
         printf("%d\n",sum);
     }
     return ;
 }