hdu 1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20353    Accepted Submission(s): 12404

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

题目大意
一个人站在'@'上，只能往'.'上走,且可以往四个方向上运动，问可以走几步，bfs搜索一下就可以了

//31ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
int n,m;
char a[22][22];
int vis[22][22];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
struct node{
    int x,y;
}pos,ans;
queue<node>que;
int bfs(int x,int y,int n,int m)
{
    memset(vis,0,sizeof(vis));
    queue<node>q;
    int cnt=1;
    pos.x=x;
    pos.y=y;
    vis[x][y]=1;
    q.push(pos);
    while(!q.empty())
    {
        pos=q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            int xx=pos.x+dir[i][0];
            int yy=pos.y+dir[i][1];
            if(xx>=0 && xx<n && yy>=0 && yy<m && a[xx][yy]=='.' && vis[xx][yy]==0)
            {
               ans.x=xx;
               ans.y=yy;
               vis[xx][yy]=1;
               cnt++;
               q.push(ans);
            }
        }
    }
    return cnt;
}
int main()
{
  while(scanf("%d%d",&m,&n)&&(n!=0 && m!=0))
  {
    int x,y;
    memset(a,0,sizeof(a));
    for(int i=0;i<n;i++)
    {
      for(int j=0;j<m;j++)
      {
        cin>>a[i][j];
        if(a[i][j]=='@')
        {
          x=i;
          y=j;
        }
      }
    }
    printf("%d
",bfs(x,y,n,m));
  }
  return 0;
}