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bfs

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;int n,m,ans;char a[100][100];int b[100][100];int fx[8]={0,0,-1,1,-1,1,-1,1},fy[8]={-1,1,0,0,-1,-1,1,1};struct tt{int a,b;};void bfs(int x,int y){b[x][y]=1;queue<tt> ss;tt o;o.a=x,o.b=y;ss.push(o);while(!ss.empty()){o=ss.front();ss.pop();for(int i=0;i<8;i++){int xx=o.a+fx[i],yy=o.b+fy[i];if(xx>=0&&yy>=0&&xx<=n&&yy<=m&&!b[xx][yy]&&a[xx][yy]=='@'){tt w;b[xx][yy]=1;w.a=xx,w.b=yy;ss.push(w);}} }}int main(){while(~scanf("%d%d",&n,&m),n){memset(b,0,sizeof(b));for(int i=0;i<n;i++)scanf("%s",a[i]);ans=0;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(!b[i][j]&&a[i][j]=='@'){ans++,bfs(i,j);}}}printf("%d\n",ans);}return 0;}

dfs

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm> using namespace std;int n,m,ans;int vis[1000][1000];char s[100][100];int fx[8]={0,0,-1,1,-1,1,-1,1},fy[8]={-1,1,0,0,-1,-1,1,1};void dfs(int x,int y){vis[x][y]=1;for(int i=0;i<8;i++){int xx=x+fx[i],yy=y+fy[i];if(xx>=0&&yy>=0&&xx<=n&&yy<=m&&!vis[xx][yy]&&s[xx][yy]=='@')dfs(xx,yy);}}int main(){while(~scanf("%d%d",&n,&m),n){memset(vis,0,sizeof(vis));for(int i=0;i<n;i++){scanf("%s",s[i]);}ans=0;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(!vis[i][j]&&s[i][j] == '@'){ans++,dfs(i,j);}}}printf("%d\n",ans);    }return 0;}

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

题意：从@开始，遇到#则不能通过，计算能到达的‘.’的个数。

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define p 100+11int n,m,ans;int a[p][p];char b[p][p];int fx[4]={0,0,-1,1},fy[4]={-1,1,0,0};void dfs(int x,int y){a[x][y]=1;for(int i=0;i<4;i++){int xx=x+fx[i],yy=y+fy[i];if(xx>=0&&yy>=0&&xx<m&&yy<n&&!a[xx][yy]&&b[xx][yy]!='#'){ans++;dfs(xx,yy);}}}int main(){int i,j,k;while(~scanf("%d%d",&n,&m),n){memset(a,0,sizeof(a));for(i=0;i<m;i++)scanf("%s",b[i]);ans=0;for(i=0;i<m;i++){for(j=0;j<n;j++){if(!a[i][j]&&b[i][j]=='@'){ans++;dfs(i,j);}}}printf("%d\n",ans);} return 0;}