Oil Deposits
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 13490 | Accepted: 7334 |
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.
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 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.
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
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
解题思路：
入门的DFS求解联通块问题，只要从图中任意一个@开始搜索就行了，对于每个走完的@，我们将其改为*，这样一直下去，总共进行了几次大的dfs,就说明有几个联通块了，什么是大的dfs呢，大的dfs就是我们说的每次从@走，走到周围（8个方向）没有@的时候。。。。在看不懂的话，就去画个图看看吧。
开始做的时候把grid的类型写成了int，改了20分钟，，，终于查出来了~~~QAQ。千万不要写错了。
代码：
# include<cstdio>
# include<iostream>
using namespace std;
# define MAX 101
char grid[MAX][MAX];
int m,n;
int next[8][2] = { {-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1} };
void dfs( int x,int y )
{
grid[x][y] = '*';
for ( int i = 0;i < 8;i++ )
{
int n_x = x+next[i][0];
int n_y = y+next[i][1];
if ( n_x<0||n_y<0||n_x>=m||n_y>=n )
continue;
if ( grid[n_x][n_y]=='@' )
dfs(n_x,n_y);
}
return;
}
int main(void)
{
int cnt;
while ( cin>>m>>n )
{
if ( m==0 )
break;
for ( int i = 0;i < m;i++ )
{
scanf("%s",grid[i]);
}
cnt = 0;
for ( int i = 0;i < m;i++ )
{
for ( int j = 0;j < n;j++ )
{
if ( grid[i][j]=='@' )
{
dfs(i,j);
cnt++;
}
}
}
cout<<cnt<<endl;
}
return 0;
}