hdu 1312 Red and Black

最新推荐文章于 2020-11-01 20:09:30 发布

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Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6461 Accepted Submission(s): 4087

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

Source

Asia 2004, Ehime (Japan), Japan Domestic

Recommend

Eddy

code:

#include<iostream>
#include<cstring>
using namespace std;
char map[1005][1005];
int vis[1005][1005];
int sx,sy;
int ans;
int dir[4][2]={0,1,0,-1,1,0,-1,0};
int  n,m;
void dfs(int sx,int sy)
{

    if (map[sx][sy]=='.'||map[sx][sy]=='@')
    {

        ans++;
    }
    else
        return ;

    for (int i=0;i<4;i++)
    {
        int nowx=dir[i][0]+sx,nowy=dir[i][1]+sy;
        if (nowx>=0&&nowy>=0&&nowy<m&&nowx<n&&map[nowx][nowy]=='.'&&vis[nowx][nowy]!=1)
        {
            //	cout<<nowx<<" "<<nowy<<endl;

            vis[nowx][nowy]=1;
            dfs(nowx,nowy);
            //	vis[nowx][nowy]=0;
        }
    }

}
int main()
{

    int sum;
    int flag;
    while (cin>>m>>n)
    {
    	if(n==0&&m==0) break;
        sum=0;
        flag=0;
        memset(map,0,sizeof(map));
        memset(vis,0,sizeof(vis));
        for (int i=0;i<n;i++)
            for (int j=0;j<m;j++)
            {
                cin>>map[i][j];
                vis[i][j]=0;
                if (map[i][j]=='@')
                {
                    flag=1;
                    sx=i;
                    sy=j;
                }
            }
        //	cout<<sx<<" "<<sy<<endl;
        if (flag==1)
        {
            ans=0;
            dfs(sx,sy);
            cout<<ans<<endl;
        }
        else
            cout<<0<<endl;
    }
    return 0;
}