hdu_1312_Red and Black

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

4559613 
比较基础的bfs题，是我学习bfs后做的第一道，初始化vis[][]数组，置所有vis的值为0，所有能访问到的位置都为1，最后输出值为1的位置的个数即可。
#include <iostream>#include <stdio.h>#include <string.h>#include <queue>using namespace std;char a[22][22];int vis[22][22];int dx[4]={-1,1,0,0};int dy[4]={0,0,-1,1};struct node{    int x,y;}cur,nex;queue<node> qu;int bfs(node sta,int w,int h){    //cout<<sta.x<<" "<<sta.y;    int cnt=0,nx,ny;    qu.push(sta);cnt++;    node p,n;    while(!qu.empty())    {        //cout<<"ok";        n=qu.front();        qu.pop();        for(int i=0;i<4;i++)        {            nx=n.x+dx[i];            ny=n.y+dy[i];            if(nx>=0 && nx<h && ny>=0&&ny<w && vis[nx][ny]==0 && a[nx][ny]=='.')            {                p.x=nx;                p.y=ny;                vis[p.x][p.y]=1;                qu.push(p);                //cout<<"ok";                cnt++;            }        }    }    return cnt;}int main(){    int w,h;    while(scanf("%d%d",&w,&h)!=EOF&&w&&h)    {        node sta;        memset(vis,0,sizeof(vis));        for(int i=0;i<h;i++)            scanf("%s",a[i]);        //把起点的位置找到        for(int i=0;i<h;i++)            for(int j=0;j<w;j++)                if(a[i][j]=='@')                {                    sta.x=i,sta.y=j;                    break;                }        vis[sta.x][sta.y]=1;        //cout<<"ok";        /*for(int i=1;i<=h;i++)        for(int j=1;j<=w;j++)        {        printf("%c ",a[i][j]);        cout<<endl;        }*/        printf("%d\n",bfs(sta,w,h));    }    return 0;}