dfs--poj1979

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Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 19476 Accepted: 10406

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) 
The end of the input is indicated by a line consisting of two zeros. 

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简单的dfs，不过要注意这个题的输入有点坑人。
#include<iostream>#include<cstring>#include<cstdio>using namespace std;int W,H;int sx,sy;int map[25][25];bool vis[25][25];int dx[]={0,0,1,-1};int dy[]={1,-1,0,0};int ans;void dfs(int x,int y){    for(int i=0;i<4;i++)    {        int tx=x+dx[i],ty=y+dy[i];        if(tx>=1&&tx<=W&&ty>=1&&ty<=H&&(!vis[tx][ty])&&(map[tx][ty]))        {            vis[tx][ty]=true;            ans++;            dfs(tx,ty);        }    }}int main(){   //freopen("in.txt","r",stdin);    char x;    while(cin>>H>>W,W||H)    {        memset(map,0,sizeof(map));        memset(vis,false,sizeof(vis));        for(int i=1; i<=W; i++)            for(int j=1; j<=H; j++)            {                cin>>x;                if(x=='.')                    map[i][j]=1;                else if(x=='@')                {                    sx=i,sy=j;                    map[i][j]=1;                }            }        ans=1;        vis[sx][sy]=1;        dfs(sx,sy);        cout<<ans<<endl;    }    return 0;}