题解报告：hdu 1312 Red and Black（简单dfs）

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

解题思路：简单深搜，水过！

AC代码：

 #include<iostream>
 using namespace std;
 const int maxn=;
 int col,row,cnt,si,sj,dir[][]={{-,},{,},{,-},{,}};char mp[maxn][maxn];
 void dfs(int x,int y){
     if(x<||y<||x>=row||y>=col||mp[x][y]=='#')return;
     mp[x][y]='#';cnt++;//标记为'#'，并且计数器加1
     for(int i=;i<;++i)
         dfs(x+dir[i][],y+dir[i][]);//直接搜索四个方向
 }
 int main(){
     while(cin>>col>>row&&(col+row)){
         for(int i=;i<row;++i){
             for(int j=;j<col;++j){
                 cin>>mp[i][j];
                 if(mp[i][j]=='@'){si=i;sj=j;}
             }
         }
         cnt=;mp[si][sj]='.';dfs(si,sj);//从'@'开始搜索
         cout<<cnt<<endl;
     }
     return ;
 }