POJ-1979

• 原题目：
  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

• 题目大意：
  给出一个w列和h行的方阵，方阵中有’.’,’#’和‘@’。其中’.’是可以移动的地方，’#’是红色的砖，是不可以移动的地 方，‘@’是最初人的地点就（图的入口），人可以上下左右移动，以0 0 为结束。
  要求输出人可以移动的砖的个数。该题利用搜索。

  注意：
  先输入w，然后输入h，但是w是列数，h是行数。

• 代码：

  #include<iostream>
  #include<cstring>
  using namespace std;
  
  void dfs(int x,int y);
  int w,h,tmp=0;
  
  char ma[100][100];
  int dx[4]={0,0,1,-1};
  int dy[4]={1,-1,0,0};
  
  int main(){
  	while(cin>>h>>w){
  		if(w==0&&h==0)  break;
  		memset(ma,'0',sizeof(ma));
  		int ax,ay;
  		for(int i=0;i<w;i++){
  			for(int j=0;j<h;j++){
  				cin>>ma[i][j];
  				if(ma[i][j]=='@'){
  					ax=i;
  					ay=j;
  				}
  			}
  		}
  		tmp=0;
  		dfs(ax,ay);
  		cout<<tmp<<endl;
  	}
  	return 0;
  }
  
  void dfs(int x,int y){
  	ma[x][y]='#';
  	tmp++;
  	for(int i=0;i<4;i++){
  		int nx=x+dx[i];
  		int ny=y+dy[i];
  		if(ma[nx][ny]=='.'&&nx>=0&&nx<w&&ny>=0&&ny<h&&ma[nx][ny]=='.'){
  			dfs(nx,ny);
  		}
  	}
  }
  Select Code
  #include<iostream>
  #include<cstring>
  using namespace std;
  
  void dfs(int x,int y);
  int w,h,tmp=0;
  
  char ma[100][100];
  int dx[4]={0,0,1,-1};
  int dy[4]={1,-1,0,0};
  
  int main(){
  	while(cin>>h>>w){
  		if(w==0&&h==0)  break;
  		memset(ma,'0',sizeof(ma));
  		int ax,ay;
  		for(int i=0;i<w;i++){
  			for(int j=0;j<h;j++){
  				cin>>ma[i][j];
  				if(ma[i][j]=='@'){
  					ax=i;
  					ay=j;
  				}
  			}
  		}
  		tmp=0;
  		dfs(ax,ay);
  		cout<<tmp<<endl;
  	}
  	return 0;
  }
  
  void dfs(int x,int y){
  	ma[x][y]='#';
  	tmp++;
  	for(int i=0;i<4;i++){
  		int nx=x+dx[i];
  		int ny=y+dy[i];
  		if(ma[nx][ny]=='.'&&nx>=0&&nx<w&&ny>=0&&ny<h&&ma[nx][ny]=='.'){
  			dfs(nx,ny);
  		}
  	}
  }

  POJ上不支持bits/stdc++的头文件，很尴尬。