POJ 1562Oil Deposits，zoj1709
Oil Deposits
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 15216 | Accepted: 8261 |
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 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
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
0122
这个题目是经典例题吧关于dfs我还是有点不熟，，在书上看到了，所以照着做了一下，在poj上accept的感觉还是不错的！
接下来是我的代码：
#include<iostream>
using namespace std;
int n,m;
const int maxn=100+5;
char s[maxn][maxn];
int dir[8][2]=
{
1,1,
-1,1,
1,0,
-1,0,
1,-1,
-1,-1,
0,-1,
0,1
};
void dfs(int x,int y)
{
s[x][y]='*';
for(int i=0;i<8;i++)
{
int tx=x+dir[i][0],ty=y+dir[i][1];
if(tx<0||tx>=n||ty<0||ty>=m)continue;
if(s[tx][ty]=='@')dfs(tx,ty);
}
return ;
}
int main()
{
while(cin>>n>>m&&m)
{
int ans=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
cin>>s[i][j];
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(s[i][j]=='@')
{
dfs(i,j);
ans++;
}
}
cout<<ans<<endl;
}
return 0;
}