hdu 1241 简单 搜索DFS （深搜）

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22567    Accepted Submission(s): 12989

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
   
   
   
   

 

Source

Mid-Central USA 1997

 

#include<stdio.h>

/**

** 深度搜索一般用递归

**/

#define Maxn 150

char s[Maxn][Maxn];//用二维数组存储图结构

int move[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}};//定义八个可行方向

int m,n,k,ans;

void dfs(int x,int y)

{

   int i;

   for(i=0;i<8;i++)//依次扫描八个方向,寻找可行点

   {

      int tx=x+move[i][0];

      int ty=y+move[i][1];

      if(tx>=0&&tx<m&&ty>=0&&ty<n&&s[tx][ty]=='@')//判断是否超出地图边界和是否可行

      {

         s[tx][ty]='*';//标记已经走过的点

         dfs(tx,ty);//以此点为起点继续深入搜索,这就是深度优先搜索

      }

   }

}

int main()

{

   int i,j;

   while(scanf("%d%d",&m,&n))

   {

      ans=0;

      if(m==0&&n==0) break;

      for(i=0;i<m;i++)

      {

         scanf("%s",s[i]);

      }

      for(i=0;i<m;i++)//分别以每个点为起点扫描一次，判断连通分量个数即可。

      {

         for(j=0;j<n;j++)

         {

            if(s[i][j]=='@')//直接忽略没有油田的点

            {

               s[i][j]='*';//标记该点已经走过

               ans++;

               dfs(i,j);

            }

         }

      }

      printf("%d\n",ans);

   }

   return 0;

}

#include<stdio.h>

/**

** 深度搜索一般用递归

**/

#define Maxn 150

char s[Maxn][Maxn];//用二维数组存储图结构

int move[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}};//定义八个可行方向

int m,n,k,ans;

void dfs(int x,int y)

{

   int i;

   for(i=0;i<8;i++)//依次扫描八个方向,寻找可行点

   {

      int tx=x+move[i][0];

      int ty=y+move[i][1];

      if(tx>=0&&tx<m&&ty>=0&&ty<n&&s[tx][ty]=='@')//判断是否超出地图边界和是否可行

      {

         s[tx][ty]='*';//标记已经走过的点

         dfs(tx,ty);//以此点为起点继续深入搜索,这就是深度优先搜索

      }

   }

}

int main()

{

   int i,j;

   while(scanf("%d%d",&m,&n))

   {

      ans=0;

      if(m==0&&n==0) break;

      for(i=0;i<m;i++)

      {

         scanf("%s",s[i]);

      }

      for(i=0;i<m;i++)//分别以每个点为起点扫描一次，判断连通分量个数即可。

      {

         for(j=0;j<n;j++)

         {

            if(s[i][j]=='@')//直接忽略没有油田的点

            {

               s[i][j]='*';//标记该点已经走过

               ans++;

               dfs(i,j);

            }

         }

      }

      printf("%d\n",ans);

   }

   return 0;

}