Poj1979 Red and Black （DFS）

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 47466		Accepted: 25523

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

Source

Japan 2004 Domestic

 

神坑的是：xy方向需要换一下

 

 #include <iostream>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 char a[][];
 int n,m;
 int res=;
 int dx[]={,-,,};
 int dy[]={,,,-};
 void dfs(int x,int y)
 {
     res++;
     a[x][y]='#';
     for(int i=;i<;i++){
         int nx=x+dx[i],ny=y+dy[i];
         if(nx>=&&nx<n&&ny>=&&ny<m&&a[nx][ny]=='.'){
             dfs(nx,ny);
         }
     }
     return ;
 }
 void solve()
 {
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(a[i][j]=='@'){
                 dfs(i,j);
             }
         }
     }
 }
 int main()
 {
     while(cin>>m>>n&&n!=&&m!=){
 
         res=;
         for(int i=;i<n;i++){
             for(int j=;j<m;j++){
                 cin>>a[i][j];
             }
         }
         solve();
         cout<<res<<endl;
     }
     return ;
 }

脱离参考书自己再根据自己的理解过一遍：

 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <string>
 #include <cstring>
 using namespace std;
 int n,m;
 char a[][];
 int sx,sy,nx,ny;
 int dx[]={,,,-};
 int dy[]={,-,,};
 int res;
 void dfs(int x,int y)
 {
     res++;
     a[x][y]='#';
     for(int i=;i<;i++){
         nx=x+dx[i],ny=y+dy[i];
         if(nx>=&&nx<m&&ny>=&&ny<n&&a[nx][ny]=='.'){
             dfs(nx,ny);
         }
     }
 }
 int main()
 {
     while(cin>>n>>m&&(n&&m)){
         for(int i=;i<m;i++){
             for(int j=;j<n;j++){
                 cin>>a[i][j];
                 if(a[i][j]=='@'){
                     sx=i,sy=j;
                 }
             }
         }
         res=;
         dfs(sx,sy);
         cout<<res<<endl;
     }
     return ;
 }