Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2101    Accepted Submission(s): 1370

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
  

 

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

 

Recommend

Eddy

来源： http://acm.hdu.edu.cn/showproblem.php?pid=1312

题目大意：@是一个man，#是红瓷砖，是不能进去的，.是可以走的地方，输入地图，输出man可以走的格子数。

解题思路：种子填充法，利用队列。先将man入队。扫描四周是否可走，可走则入队，出队中心点。依次出队入队。直到队空。

总结：注意坐标的统一。初始化flag时，当是二维的时候不能用mamset（flag，0，sizeof（int））。

AC代码：

#include<iostream>
#include <queue>
using namespace std;
char map[21][21];
int di[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
typedef struct
{
 int x,y;
}point;
point man;
int n,m;
void tiles()
{
 int flag[21][21]={0};
 int i,c=0;
 queue<point> q;
 point t,t1;
 q.push(man);
 flag[man.y][man.x]=1;
 c++;
 while(!q.empty())
 {
  t=q.front();
  q.pop();
  for(i=0;i<4;i++)
  {
   t1.x=t.x+di[i][0];
   t1.y=t.y+di[i][1];
   if(map[t1.y][t1.x]!='#'&&flag[t1.y][t1.x]==0&&(t1.x<n&&t1.x>=0)&&(t1.y<m&&t1.y>=0))
   {
    q.push(t1);
    flag[t1.y][t1.x]=1;
    c++;
   }
  }
 }
 printf("%d/n",c);
}
int main()
{
 int i,j;
 while(scanf("%d%d",&n,&m))
 {
  getchar();
  if(n==0&&m==0) break;
  for(i=0;i<m;i++)
  {
   for(j=0;j<n;j++)
   {
    scanf("%c",&map[i][j]);
    if(map[i][j]=='@')
    {
     man.x=j;man.y=i;
    }
   }
   getchar();
  }
  tiles();
 }
 return 0;
}