HDU1241-Oil Deposits

Oil Deposits
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 9   Accepted Submission(s) : 6
Problem DescriptionThe 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.
 

InputThe 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.
 

OutputFor 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 Input1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0
 

Sample Output0
1
2
2

题意：所给你一个地图，有两种元素分别为‘.’与@,然后要求你找出不相连的@的个数，@的邻接与@的对角都算是相连的。

注意：第三组测试数据的"5 5“后有个空格，复制粘贴会出错，建议手动输入
//DFS

#include<stdio.h>
#include<string.h>
struct node
{
   char V;
   int flag;
}a[200][200];
int n,m;
void Fun(int x,int y)
{
     if(x+1<n&&y-1>=0&&a[x+1][y-1].flag==0&&a[x+1][y-1].V=='@')
     {
        a[x+1][y-1].flag=1;
        Fun(x+1,y-1);
     }

     if(x+1<n&&y+1<m&&a[x+1][y+1].flag==0&&a[x+1][y+1].V=='@')
     {
        a[x+1][y+1].flag=1;
        Fun(x+1,y+1);
     }

     if(x-1>=0&&y-1>=0&&a[x-1][y-1].flag==0&&a[x-1][y-1].V=='@')
     {
        a[x-1][y-1].flag=1;
        Fun(x-1,y-1);
     }

     if(x-1>=0&&y+1<m&&a[x-1][y+1].flag==0&&a[x-1][y+1].V=='@')
     {
        a[x-1][y+1].flag=1;
        Fun(x-1,y+1);
     }

     if(x+1<n&&a[x+1][y].V=='@'&&a[x+1][y].flag==0)
     {
        a[x+1][y].flag=1;
        Fun(x+1,y);
     }

     if(x-1>=0&&a[x-1][y].V=='@'&&a[x-1][y].flag==0)
     {
        a[x-1][y].flag=1;
        Fun(x-1,y);
     }

     if(y+1<m&&a[x][y+1].V=='@'&&a[x][y+1].flag==0)
     {
        a[x][y+1].flag=1;
        Fun(x,y+1);
     }

     if(y-1>=0&&a[x][y-1].V=='@'&&a[x][y-1].flag==0)
     {
        a[x][y-1].flag=1;
        Fun(x,y-1);
     }

     return;
}
int main()
{
    int i,j,sum;
    while(scanf("%d %d",&n,&m),n!=0&&m!=0)
    {
       memset(a,0,sizeof(a));
       for(i=0;i<n;i++)
       {
          getchar();
          for(j=0;j<m;j++)
          scanf("%c",&a[i][j].V);
       }
        sum=0;
        for(i=0;i<n;i++)
        {
           for(j=0;j<m;j++)
           {
               if(a[i][j].V=='@'&&a[i][j].flag==0)
               {
                  a[i][j].flag=1;
                  Fun(i,j);
                  sum++;
               }
           }
        }
        printf("%d\n",sum);
    }
    return 0;
}