杭电1312 Red and Black

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

 

题意是：从@（本身算一个）开始找能够走通的‘。’有几个

思路：利用广度搜索法，先找其四周的位置是否为‘。’，如果是就先将他们的位置记录在数组中，并将该位置的记为‘#’，直到找完为止。

代码：

#include"iostream"
#define N 21
using namespace std;

typedef struct
{int x;
 int y;
}node;

node jilu[N*N];
char map[N][N];   //记录该图；
int s1,s2;        //记录起始位置；
int s;          //记录个数，起始点为1;
int star,end;
int fx[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int h,w;

void BB()
{
 s=1;
 star=0;
 end=1;
    node first,next;
    jilu[0].x=s1;
    jilu[0].y=s2;
 map[s1][s2] = '#';//本身就是一个可以的点（即s=1），要将它去掉（防止重复）
 while(end!=star)
 {
  first=jilu[star];
  star++;           //去掉头个；
  for(int i=0;i<4;i++)   //判断四个方位是否满足；
  {
   next.x=first.x+fx[i][0];
   next.y=first.y+fx[i][1];
   if(next.x>=0 && next.x<h && next.y>=0 && next.y<w && map[next.x][next.y]!='#')
   {
    s++;
    map[next.x][next.y] = '#';
    jilu[end]=next;
       end++;
   } 
  }
 }
 cout<<s<<endl;
}

int main()
{
 while(cin>>w>>h&&w&&h)
   {
 for(int j=0;j<h;j++ )
 {
  for(int k=0;k<w;k++)
  {
   cin>>map[j][k];
   if(map[j][k]=='@')
    s1=j,s2=k;
  }
 
 }
 BB();
   }
 return 0;

}