Oil Deposits（poj 1526 DFS入门题）

http://poj.org/problem?id=1562

                                                                                                   Oil Deposits

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12595		Accepted: 6868

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

0
1
2
2
题目大意：如果'@'周围的8个方向有'@'就认为它们是同一油田。求总油田的个数。
本题是dfs的入门题，我的dfs只能用很烂来形容，dfs利用了栈的思想，先进后出。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define N 1000001
using namespace std;
int n,m;
char map[101][101];
int v[101][101];
int jx[]={1,1,1,-1,-1,-1,0,0};
int jy[]={-1,0,1,-1,0,1,-1,1};
void dfs(int x,int y)
{
    v[x][y]=1;
    int tx,ty;
    for(int i=0;i<8;i++)
    {
       tx=x+jx[i];
       ty=y+jy[i];
       if(tx>=0&&tx<n&&ty>=0&&ty<m&&map[tx][ty]=='@'&&v[tx][ty]==0)
       {
           dfs(tx,ty);
       }
    }
}
int main()
{
    int ans=0;
    while(scanf("%d%d",&n,&m)!=EOF&&n!=0)
    {
        ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%*c%s",map[i]);
        }
        memset(v,0,sizeof(v));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(map[i][j]=='@'&&v[i][j]==0)
                {
                    dfs(i,j);
                    ans++;
                }

            }
        }
        printf("%d
",ans);

    }
    return 0;
}