Red and Black

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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

算法：DFS

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int x,y,xx,yy;int w,h;char str[25][25];int vis[25][25];int ans;int f1[4]={0,0,-1,1},f2[4]={-1,1,0,0};void dfs(int x,int y){    vis[x][y] = 1,ans++;    for(int i = 0; i < 4; i++){        int xx = x + f1[i],yy = y + f2[i];        if(xx >= 0 && yy >= 0 && xx < h && yy < w && !vis[xx][yy] && str[xx][yy] == '.')            dfs(xx,yy);    }}int main(){while(~scanf("%d %d",&w,&h)&&(w!=0||h!=0)){memset(vis,0,sizeof(vis));for(int i = 0 ; i < h ; i++)scanf("%s",str[i]);for(int i = 0 ; i < h ; i++)for(int j = 0 ; j < w ; j++)if(str[i][j]=='@')x=i,y=j;ans=0;dfs(x,y);printf("%d\n",ans);}return 0;}

BFS

#include<cstdio>#include<algorithm>#include<cstring>#include<queue>using namespace std;int x,y,xx,yy;int w,h;char str[25][25];int vis[25][25];int ans;int f1[4]={0,0,-1,1},f2[4]={-1,1,0,0};struct node{int x,y;};void bfs(int x , int y){vis[x][y]=1;node o;o.x=x,o.y=y;queue <node> q;q.push(o);while(!q.empty()){o=q.front();ans++;q.pop();for(int i = 0; i < 4; i++){            int xx = o.x + f1[i],yy = o.y + f2[i];            if(xx >= 0 && yy >= 0 && xx < h && yy < w && !vis[xx][yy] && str[xx][yy] == '.'){                node w;                vis[xx][yy] = 1;                w.x = xx,w.y = yy;                q.push(w);            }        }}}int main(){while(~scanf("%d %d",&w,&h)&&(w!=0||h!=0)){memset(vis,0,sizeof(vis));for(int i = 0 ; i < h ; i++)scanf("%s",str[i]);for(int i = 0 ; i < h ; i++)for(int j = 0 ; j < w ; j++)if(str[i][j]=='@')x=i,y=j;ans=0;bfs(x,y);printf("%d\n",ans);}return 0;}