HDU 1312 Red and Black

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4306    Accepted Submission(s): 2798

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
  
  

 

思路：简单的搜索题，从‘@’出发，只能站在‘.’上，不能站在‘#’上，求能踩到的.的个数（包括起始点@）

我用BFS做

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int map[22][22];
char a[22][22];
int n,m,fx,fy,kind;
queue<int> q;
void Init()
{
	int i;
  for(i=0;i<=n;i++) {a[i][0]='#';a[i][m+1]='#';}
  for(i=0;i<=m;i++) {a[0][i]='#';a[n+1][i]='#';}
  memset(map,0,sizeof(map));
}
void bfs(int x,int y)
{
   map[x][y]=++kind;
   q.push(x),q.push(y);
   while(!q.empty())
   {
	   x=q.front();q.pop();
	   y=q.front();q.pop();
      if(a[x-1][y]=='.'&&!map[x-1][y])
	  {
	     map[x-1][y]=++kind;
		 q.push(x-1),q.push(y);
	  }
      if(a[x+1][y]=='.'&&!map[x+1][y])
	  {
	     map[x+1][y]=++kind;
		 q.push(x+1),q.push(y);
	  }
	  if(a[x][y-1]=='.'&&!map[x][y-1])
	  { 
	     map[x][y-1]=++kind;
		 q.push(x),q.push(y-1);
	  }
	  if(a[x][y+1]=='.'&&!map[x][y+1])
	  {
	     map[x][y+1]=++kind;
		 q.push(x),q.push(y+1);
	  }
   }
}
int main()
{
   int i,j;
   while(scanf("%d%d",&m,&n)!=EOF&&(n+m))
   {
	   Init();
      for(i=1;i<=n;i++)
	  {
		  getchar();
		  for(j=1;j<=m;j++)  
		  {
			  scanf("%c",&a[i][j]);
			  if(a[i][j]=='@')
			  {
			     fx=i;fy=j;
			  }
		  }
	  }
	  kind=0;
	  bfs(fx,fy);
      printf("%d\n",kind);
   }
   return 0;
}