HDU 1241 Oil Deposits（深搜）

题意:

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

Simple Input:

1
1

*

3
5

*@*@*

**@**

*@*@*

1 8

@@****@*

5 5

****@

*@@*@

*@**@

@@@*@

@@**@

0 0

Output:

0

1

2

2

题目分析:

本题是要求出样例描绘的图中有多少油田（独立的），即8个方向中不能有任何交集的为两块独立的油田，，因此可以用dfs算法做出.

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char map[1000][1000];
int n,m;
int dir[8][2]={{1,0},{1,-1},{1,1},{0,-1},{0,1},{-1,0},{-1,-1},{-1,1}};//定义8个方向
void dfs(int x,int y)
{
int x1,y1;
map[x][y]='*';
for(int i=0;i<8;i++)
{
x1=x+dir[i][0];
y1=y+dir[i][1];
if(x1<0||x1>=n||y1<0||y1>=m)
continue;
if(map[x1][y1]=='@')
{
dfs(x1,y1);
}
}
}
int main()
{
int count;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(m==0)
break;
getchar();
for(int i=0;i<n;i++)
{
scanf("%s",map[i]);
}
count=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(map[i][j]=='@')//找到油田，并来确定范围
{
dfs(i,j);
count++;
}
}
}
printf("%d\n",count);
}
return 0;
}