dfs水题hdu1241—是什么让一个植物人从b站中惊醒

是这该死的生活，要恰饭的嘛

Oil Deposits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 48272    Accepted Submission(s): 27768   Problem Description 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 0 1 2 2 离超越马爸爸又近了一步的AC代码： #include #include #include using namespace std; vectorch; int idx[105][105]; int m,n; void dfs(int r,int c,int id) {     if(r<0||r>=m||c<0||c>=n)         return;     if(idx[r][c]>0||ch[r][c]!='@')         return ;     idx[r][c]=id;     for(int pr=-1;pr<=1;pr++)         for(int pc=-1;pc<=1;pc++){             if(pc||pr)                 dfs(r+pr,c+pc,id);         } }    int main() {     int key;     string word;     while((cin>>m>>n)&&m&&n){         key=0;         ch.clear();         for(int i=0;i             cin>>word;                 ch.push_back(word);             fill(idx[i],idx[i]+n,0);         }         for(int i=0;i             for(int j=0;j                 if(ch[i][j]=='@'&&idx[i][j]==0)                     dfs(i,j,++key);             }         }         cout<     }     return 0; }