C

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid. 

Input

The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket. 

Output

For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets. 

Sample Input

1 1*3 5*@*@***@***@*@*1 8@@****@*5 5 ****@*@@*@*@**@@@@*@@@**@0 0 

Sample Output

0122

题意：问有几个“@”的联通块。。。。

思路：dfs直接搜索好了。。。当找到一个“@”符号时，直接以他为原点向周围8个方向搜索，如果还有的话再以新找到的“@”为原点找，当找不到时就返回回溯，再重新找一个“@”继续搜索就好了；用一个变量记录次数。。。

下面附上代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;int sx,sy,nx,ny,ans,m,n;char mapp[105][105];int mx[8]={0,0,1,-1,1,-1,1,-1};int my[8]={1,-1,0,0,1,-1,-1,1};int vis[205][205];bool check(int x,int y){if(x<0||y<0||x>=n||y>=m||vis[x][y])return false;return true;}void dfs(int x,int y){vis[x][y]=1;for(int i=0;i<8;i++){nx=x+mx[i];ny=y+my[i];if(!check(nx,ny))continue;else if(mapp[nx][ny]=='@')dfs(nx,ny);}}int main(){while(~scanf("%d %d",&n,&m)&&(n||m)){ans=0;memset(vis,0,sizeof(vis));for(int i=0;i<n;i++)scanf("%s",mapp[i]);for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(mapp[i][j]=='@'&&!vis[i][j]){ans++;dfs(i,j);}}}printf("%d\n",ans);}return 0;}