• 原題目：
  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
  .#.........
  .#.#######.
  .#.#.....#.
  .#.#.###.#.
  .#.#[email protected]
   
  #.#.
  .#.#####.#.
  .#.......#.
  .#########.
  ...........
  11 6
  ..#..#..#..
  ..#..#..#..
  ..#..#..###
  ..#..#..#@.
  ..#..#..#..
  ..#..#..#..
  7 7
  ..#.#..
  ..#.#..
  ###.###
  [email protected]
  ###.###
  ..#.#..
  ..#.#..
  0 0
  

  Sample Output

  45
  59
  6
  13
  

• 題目大意：
  給出一個w列和h行的方陣，方陣中有’.’,’#‘和’@’。其中’.‘是可以移動的地方，’#‘是紅色的磚，是不可以移動的地 方，’@'是最初人的地點就（圖的入口），人可以上下左右移動，以0 0 為結束。
  要求輸出人可以移動的磚的個數。該題利用搜索。

注意：
先輸入w，然後輸入h，但是w是列數，h是行數。

• 程式碼：

#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上不支援bits/stdc++的標頭檔案，很尷尬。