HDU 1312 Red and Black （dfs）

Posted 2020-08-12 风中的簌雨

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

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#[email protected]#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7 7

..#.#..

..#.#..

###.###

[email protected]

###.###

..#.#..

..#.#..

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：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <stack>
 6 #include <queue>
 7 using namespace std;
 8 int w,h;
 9 int a[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};  //位置数组 
10 char str[22][22];
11 int f[22][22];  //标记该位置是否走过  
12 int sum;
13 void dfs(int x,int y)
14 {
15     f[x][y] = 1;
16     for (int i = 0; i < 4; i ++)
17     {
18         int x1=x + a[i][0];
19         int y1=y + a[i][1];
20         if (x1 >= 0 && x1 < h && y1 >= 0 && y1 < w && str[x1][y1]!=‘#‘ && f[x1][y1] == 0) 
21         {    //判断边界、是否是不能经过的红地板、该地板是否已经经过 
22             sum ++;
23             dfs(x1,y1);
24         }
25     }
26 }
27 int main ()
28 {
29     int i,j,x,y;
30     while (scanf("%d%d",&w,&h),w&&h)
31     {
32         for (i = 0; i < h; i ++)
33             scanf("%s",str[i]);
34 
35         memset(f,0,sizeof(f));
36         for (i = 0; i < h; i ++)
37         for (j = 0; j < w; j ++)
38             if (str[i][j] == ‘@‘) //找出人的位置 
39             {
40                 x = i;
41                 y = j;
42                 break;
43             }
44         sum = 1;
45         dfs(x,y);
46         printf("%d\n",sum);
47     }
48     return 0;
49 }