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): 10133    Accepted Submission(s): 6321

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

....#.

.....#

......

......

......

......

......

#@...#

.#..#.

0 0

 

 

Sample Output

45

 

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

 

PS:简单题

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<queue>
using namespace std;
int n,m;
int xi,yj;
char map[25][25];
int vis[25][25];
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//上下左右
struct node{
    int x;
    int y;
};

bool check(int x,int y)
{
    if(x>=1&&x<=n && y>=1&&y<=m && map[x][y]=='.' && !vis[x][y])
    return true;
    return false;
}

int bfs()
{
    memset(vis,0,sizeof(vis));
    int i,sum=1;
    node p,q;
    queue<struct node>Q;
    p.x=xi;
    p.y=yj;
    Q.push(p);
    while(!Q.empty())
    {
        p=Q.front();
        Q.pop();

        for(i=0;i<4;i++)
        {
            q.x=p.x+dir[i][0];
            q.y=p.y+dir[i][1];
            if(check(q.x,q.y))
            {
                vis[q.x][q.y]=1;
                sum++;
                Q.push(q);
            }
        }
    }
    return sum;
}


int main()
{
    int i,j;
    int count;
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&m,&n))
    {

        if(!n&&!m)
        break;
        for(i=1;i<=n;i++)
        {
            getchar();
            for(j=1;j<=m;j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j]=='@')
                {
                    xi=i;
                    yj=j;
                }
            }
        }
        count=bfs();
        printf("%d
",count);
    }
    return 0;
}