ACM----HDU-1241 dfs2

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

解题思路：两个相邻的含油油井属于同一个油井，想要解出区域内有多少个不同的油井。只需在碰到含油的油井时，遍历它周围的八个油井，将有油的点归零，之后将计数器加一，遍历完所有有油的油井后，输出计数器的值即为所求。

#include<cstdio>
#include<iostream>
using namespace std;
char map[200][200];
int num=0;
int m=0,n=0;
int xc[8]={1,1,0,-1,-1,-1,0,1};
int yc[8]={0,-1,-1,-1,0,1,1,1};
void dfs(int x, int y)
{
    map[x][y] = '*';
    for(int i = 0; i < 8; i++){//寻找该油井周围的八个油井
        int vx=x+xc[i];
        int vy=y+yc[i];
        if (map[vx][vy] == '@' && vx < m && vx >= 0 && vy < n && vy >= 0) dfs(vx, vy);
    }
}
int main(){

    while(scanf("%d%d",&m,&n)!=EOF&&m!=0&&n!=0){
        num=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                cin>>map[i][j];
            }
        }
      for(int i=0;i<m;i++){
        for(int j=0;j<n;j++){
            if(map[i][j]=='@'){
                dfs(i,j);num++;
            }
        }
    }
        printf("%d\n",num);
    }
   return 0;
}