hdu 1312 Red and Black (dfs+bfs)

Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15731 Accepted Submission(s): 9730


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

dfs

#include<stdio.h>
#include<string.h>
#include <string>
#include <queue>
#include <iostream>
using namespace std;
char map[103][103];
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int m,n,cnt;
/*void dfs(int x,int y)
{
    if (map[x][y]=='#')
        return ;
    map[x][y]='#';
    cnt++;
    for (int i=0;i<4;i++)
      {
          int ax=x+dir[i][0];
            int ay=y+dir[i][1];
            if (ax<m&&ax>=0&&ay<n&ay>=0)
                dfs(ax,ay);
      }
}*/
void dfs(int x,int y)
{
    int i;
    for (i=0;i<4;i++)
    {
        int ax=x+dir[i][0];
        int ay=y+dir[i][1];
        if (ax<m&&ax>=0&&ay<n&&ay>=0&&map[ax][ay]!='#')
        {
            cnt++;
            map[ax][ay]='#';
            dfs(ax,ay);
        }
    }
}
int main()
{
    int i,j,px,py;
    while (~scanf("%d%d",&n,&m),m+n)
    {
        getchar();
            for (i=0;i<m;i++)
            {
                for (j=0;j<n;j++)
                {
                     scanf("%c",&map[i][j]);
                     if (map[i][j]=='@')
                     {
                         px=i;
                         py=j;
                     }
                }
                getchar();
            }
            cnt=1;
            map[px][py]='#';
            dfs(px,py);
            printf("%d\n",cnt);
    }
    return 0;
}

bfs

#include<stdio.h>
#include<string.h>
#include <string>
#include <queue>
#include <iostream>
using namespace std;
char map[103][103];
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int m,n,cnt;
typedef struct node{
    int x,y;
}node;
void bfs(int x,int y)
{
    queue<node> Q;
    node start,t1,t2;
    start.x=x,start.y=y;
    Q.push(start);
    while (!Q.empty())
    {
        t1=Q.front();
        Q.pop();
        for (int i=0;i<4;i++)
          {
              int ax=t1.x+dir[i][0];
              int ay=t1.y+dir[i][1];
              if (ax<m&&ax>=0&&ay<n&&ay>=0&&map[ax][ay]!='#')
              {
                  t2.x=ax,t2.y=ay;
                  map[ax][ay]='#';
                  cnt++;
                  Q.push(t2);
              }
          }
    }
}
int main()
{
    int i,j,px,py;
    node s;
    while (~scanf("%d%d",&n,&m),m+n)
    {
        getchar();
            for (i=0;i<m;i++)
            {
                for (j=0;j<n;j++)
                {
                     scanf("%c",&map[i][j]);
                     if (map[i][j]=='@')
                     {
                         px=i;
                         py=j;
                         map[px][py]='#';
                     }
                }
                getchar();
            }
            cnt=1;
            bfs(px,py);
            printf("%d\n",cnt);
    }
    return 0;
}