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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).

代码：

#include<iostream>#include<cstring>#include <algorithm>#include<queue>#include<vector># define Maxn 110#define MP(a, b) make_pair(a, b)using namespace std;typedef pair<int, int> pii;char mp[Maxn][Maxn];int vis[Maxn][Maxn];int cou,m,n;int dx[]={0,0,1,-1};int dy[]={-1,1,0,0};int judge(int x,int y){    return x>=0&&y>=0&&x<n&&y<m;}void dfs(int x,int y){    vis[x][y]=1;    for(int i=0;i<4;i++){        int tx=x+dx[i],ty=y+dy[i];        if(!vis[tx][ty]&&judge(tx,ty)&&mp[tx][ty]=='.')            {cou++;dfs(tx,ty);}    }}int main(){    int x,y;    while(cin>>m>>n,m){        memset(vis,0,sizeof vis);        for(int i=0;i<n;i++)            for(int j=0;j<m;j++){                cin>>mp[i][j];                if(mp[i][j]=='@')                {x=i;y=j;}            }        cou=0;        dfs(x,y);        cout<<cou+1<<endl;    }    return 0;}

注：这题相当于找当前连通分支的元素个数。