ＨＤＵ　１２４１

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

 

這題還是比較好理解的，有著重疊的子問題，呼叫遞迴即可，感覺和ＤＰ有點不同，

ＤＰ重疊子問題後會有狀態的轉移，這個只用遞迴即可。（個人理解，剛學，比較菜）

 1 #include<bits/stdc++.h>
 2 using namespace std ;
 3 
 4 char a[105][105];
 5 void
 
 DFS(int x, int y)
 6 {
 7     a[x][y]=0;
 8     int X[8]={-1,-1,0,1,1,1,0,-1};
 9     int Y[8]={0,1,1,1,0,-1,-1,-1};
10     for(int k=0; k<8; k++){
11         int i=x+X[k];
12         int j=y+Y[k];
13         if(a[i][j]=='@'){
14             DFS(i,j);
15         }
16     }
17 }
18 int main ()
19 {
20     int
 
 n,m;
21     while(cin>>n>>m)
22     {
23         if(n==0&&m==0)
24             break;
25 
26         memset(a, 0, sizeof(a));
27         for(int i=0; i<n; i++)
28             for(int j=0; j<m; j++)
29                 cin>>a[i][j];
30 
31         int cnt=0;
32         for(int i=0; i<n; i++)
33             for(int j=0; j<m; j++)
34              if(a[i][j]=='@'){
35                 cnt++;
36                 DFS(i,j);
37             }
38         cout<<cnt<<endl;
39     }
40     return 0;
41 }