ZOJ 1709 Oil Deposits(油田问题DFS)

Oil Deposits

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

#include<iostream>

#include<stdio.h>

using namespace std;

const int maxn=105;

char grid[maxn][maxn];

int dir[8][2]={{-1,1},{0,1},{1,1},{-1,0},{1,0},{-1,-1},{0,-1},{1,-1}};

int m,n;//之所以定义为全局变量是因为在dfs函数中要用到

int main()

{

void dfs(int,int);

while(scanf("%d%d",&m,&n)!=EOF)

{

int i,j;

if(m==0){break;}

for(i=0;i<m;i++)

{scanf("%s",&grid[i]);}//用字符串而不用双重for循环进行输入避免了字符中的空格和回车问题

int sum=0;

for(i=0;i<n;i++)

for(j=0;j<m;j++)

{

if(grid[i][j]=='@')

{

dfs(i,j);

sum++;

}

}

printf("%d\n",sum);

}

return 0;

}

void dfs(int x,int y)

{

grid[x][y]='*';

for(int i=0;i<8;i++)

{

int xi=x+dir[i][0];

int yi=y+dir[i][1];

if(xi<0||xi>=n||yi<0||yi>=m){continue;}

/*注意这里的x和y是所在位置的行数和列数，不是所在位置的横坐标和纵坐标*/

if(grid[xi][yi]=='@'){dfs(xi,yi);}

}

}