HDU1241 Oil Deposits —— DFS求连通块

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1241

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33688    Accepted Submission(s): 19585

Problem 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 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

 

Source

Mid-Central USA 1997

题解：

典型的dfs求连通块问题。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 100+10;

char M[MAXN][MAXN];
int vis[MAXN][MAXN];
int n, m, dir[8][2] = {0,1,1,0,0,-1,-1,0, 1,1,-1,1,1,-1,-1,-1};

void dfs(int x, int y)
{
    vis[x][y] = 1;
    for(int i = 0; i<8; i++)
    {
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if( xx>=1 && xx<=n && yy>=1 && yy<=m && M[xx][yy]=='@' && !vis[xx][yy])
            dfs(xx,yy);
    }
}

int main()
{
    while(scanf("%d%d",&n,&m) && m )
    {
        for(int i = 1; i<=n; i++)
            scanf("%s", M[i]+1);

        int ans = 0;
        ms(vis, 0);
        for(int i = 1; i<=n; i++)
        for(int j = 1; j<=m; j++)
        {
            if(M[i][j]=='@' && !vis[i][j])
            {
                ans++;
                dfs(i,j);
            }
        }
        printf("%d
", ans);
    }

}