HDU1241-经典dfs

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22306 Accepted Submission(s): 12860

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

题意：
在m*n的区域内有用@表示的油田，相连的@认为是一个油田，相连是指上、下、左、右、东北、西北、东南、西南，一共8个方向。问有多少块油田。

思路：
很明显是dfs，由于是求有多少块，那么，把‘@’访问后变成 ‘ * ’，对每个是@的点dfs即可

#include <iostream>#include <string.h>#include <cstdio>int m, n;const int maxn = 1000+10;char arr[maxn][maxn];int dir[8][2] = {{-1,0},{-1,-1},{-1,1},{0,1},{0,-1},{1,1},{1,0},{1,-1}};using namespace std;void DFS(int x, int y){    for(int i=0;i<8;i++)    {        int ddx = dir[i][0] + x;        int ddy = dir[i][1] + y;        if(0<=ddx && ddx<m && 0<=ddy && ddy<n && arr[ddx][ddy]=='@')        {            arr[ddx][ddy] = '*';            DFS(ddx, ddy);        }    }}int main(){    //freopen("input.txt","r",stdin);    //freopen("output.txt","w",stdout);   while(scanf("%d%d",&m,&n))   {      int ans=0;      if(m==0 && n==0) break;      memset(arr,0,sizeof(arr));      for(int i=0; i<m;i++)          for(int j=0; j<n; j++)              cin>>arr[i][j];      for(int i=0; i<m; i++)         for(int j=0; j<n; j++){            if(arr[i][j] == '@'){               arr[i][j] = '*';               ans++;               DFS(i, j);            }         }        cout<<ans<<endl;   }   return 0;}

如有不对欢迎指出