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=25;
int col,row,cnt,si,sj,dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};char mp[maxn][maxn];
void dfs(int x,int y){
    if(x<0||y<0||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=0;i<4;++i)
        dfs(x+dir[i][0],y+dir[i][1]);//直接搜索四个方向
}
int main(){
    while(cin>>col>>row&&(col+row)){
        for(int i=0;i<row;++i){
            for(int j=0;j<col;++j){
                cin>>mp[i][j];
                if(mp[i][j]=='@'){si=i;sj=j;}
            }
        }
        cnt=0;mp[si][sj]='.';dfs(si,sj);//从'@'开始搜索
        cout<<cnt<<endl;
    }
    return 0;
}