hdu 1312 Red And Black

hdu 1312 的传送门

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).
解题思路：最简单的dfs和bfs，个人还是建议用dfs，时间短，而且还好写；
具体的上代码；

#include <iostream>#include <cstring>using namespace std;char map[30][30];bool vis[30][30];int qx[900],qy[900];//bfsint dir[4][2]= {1,0,-1,0,0,1,0,-1};int n,m,ans;/*void bfs(int x, int y){    int l=0,r=0;    qx[r]=x,qy[r++]=y;    vis[x][y]=1;    ans++;    int nx,ny;    while(l<r)    {        int curx=qx[l],cury=qy[l++];        for(int i=0; i<4; i++)        {            nx=curx+dir[i][0];            ny=cury+dir[i][1];            if(nx>=0&&nx<n && ny>=0&&ny<m && !vis[nx][ny] && map[nx][ny]!='#')            {                vis[nx][ny]=1;                ans++;                qx[r]=nx,qy[r++]=ny;            }        }    }}*/void dfs(int x, int y){    ans++;    vis[x][y]=1;//可以不用    map[x][y]='#';    for(int i=0; i<4; i++)    {        int nx=x+dir[i][0];        int ny=y+dir[i][1];        if(nx>=0&&nx<n && ny>=0&&ny<m &&!vis[nx][ny]&& map[nx][ny]!='#')        {            dfs(nx, ny);        }    }    //cout<<ans<<endl;}int main(){    while(cin>>m>>n, m,n)    {        memset(vis, 0, sizeof(vis));        ans=0;        int sx,sy;        for(int i=0; i<n; i++)        {            for(int j=0; j<m; j++)            {                cin>>map[i][j];                if(map[i][j] == '@')                sx=i,sy=j;            }        }        //bfs(sx, sy);        dfs(sx, sy);        cout<<ans<<endl;    }    return 0;}/*6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0*/