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

45

59

6

13

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

 #include<bits/stdc++.h>
 using namespace std;
 int w,h;
 char s[][];
 int maxx=;
 int sx,sy;
 int dirx[]={,-,,},diry[]={,,,-};
 void dfs(int x,int y)
 {
     for(int i=;i<;i++)
     {
         int xx=x+dirx[i];
         int yy=y+diry[i];
         if(xx<||yy<||xx>=h||yy>=w)continue;
         if(s[xx][yy]=='.')
         {
             s[xx][yy]='#';
             maxx++;
             dfs(xx,yy);
         }
     }
 }
 int main()
 {
     while(~scanf("%d %d",&w,&h),w+h)
     {
         maxx=;
         memset(s,'\0',sizeof(s));
         for(int i=;i<h;i++)
         {
             getchar();
             for(int j=;j<w;j++)
             {
                 scanf("%c",&s[i][j]);
                 if(s[i][j]=='@')
                 {
                     sx=i;
                     sy=j;
                 }
             }
         }
         dfs(sx,sy);
         printf("%d\n",maxx+);//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
     }
     return ;
 }