hdu1312Red and Black(迷宫dfs，一遍)

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26802 Accepted Submission(s): 16176

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

题意：给出一个类似迷宫的东西，#表示红砖，@表示起点，也是黑砖，点号表示黑砖，每次只能走黑砖，问最多经过多少块黑砖。每次只能走上下左右四个方向

题解：不用剪枝的dfs.....感动.jpg。就直接看下一个地方能不能走，能走就标记并计数。只要走一遍，所以不需要回溯。因为写的是看下一步能不能走，所以没有算起点，答案就要加1

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int w,h;
 4 char s[25][25];
 5 int maxx=0;
 6 int sx,sy;
 7 int dirx[4]={1,-1,0,0},diry[4]={0,0,1,-1};
 8 void dfs(int x,int y)
 9 {
10     for(int i=0;i<4;i++)
11     {
12         int xx=x+dirx[i];
13         int yy=y+diry[i];
14         if(xx<0||yy<0||xx>=h||yy>=w)continue;
15         if(s[xx][yy]=='.')
16         {
17             s[xx][yy]='#';
18             maxx++;
19             dfs(xx,yy);
20         }
21     }
22 }
23 int main()
24 {
25     while(~scanf("%d %d",&w,&h),w+h)
26     {
27         maxx=0;
28         memset(s,'',sizeof(s));
29         for(int i=0;i<h;i++)
30         {
31             getchar();
32             for(int j=0;j<w;j++)
33             {
34                 scanf("%c",&s[i][j]);
35                 if(s[i][j]=='@')
36                 {
37                     sx=i;
38                     sy=j;
39                 }
40             }
41         }
42         dfs(sx,sy);
43         printf("%d
",maxx+1);//最后答案加1是因为我每次判断下一个是可走的，就+1，这样就没有把起点计算进去 
44     }
45     return 0;
46 }