hdoj problem 1241 Oil Deposits(深搜 DFS)

Oil Deposits

http://acm.hdu.edu.cn/showproblem.php?pid=1241

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13110    Accepted Submission(s): 7593

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

0122

 

Source

Mid-Central USA 1997

 

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/*

这是我第一次做出来的！

此题与hdoj 2952 做法 一模一样

http://acm.hdu.edu.cn/showproblem.php?pid=2952

*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char map[105][105];
int x[8]={0,0,-1,1,1,1,-1,-1};//深搜的关键之处
int y[8]={1,-1,0,0,-1,1,1,-1};
int m,n;
void DFS(int xx,int yy)
{
     int i,a,b;
     for(i=0;i<8;i++)八个方向
     {
      int a=x[i]+xx;
      int b=y[i]+yy;
      if(a>=0&&b>=0&&a<m&&b<n&&map[a][b]=='@')//递归的边界
      {
          map[a][b]='*';//已经遍历的直接标识一下
        DFS(a,b);
      }
    }
}


int main()
{
  while(scanf("%d%d",&m,&n)&&(n||m))
  {
      int i,j;
      getchar();
      for(i=0;i<m;i++)
        scanf("%s",map[i]);
      
        int count=0;
        for(i=0;i<m;i++)
          for(j=0;j<n;j++)
            if(map[i][j]=='@')//一个一个访问
            {
            count++;
          DFS(i,j);    
          }
          printf("%d\n",count);
  }
  return 0;    
}