UVA 572 Oil Deposits

最新推荐文章于 2021-05-02 17:02:24 发布

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本文介绍了一个用于探测矩形区域地下油藏的算法。通过输入由字符‘@’(表示油藏)和‘*’(表示非油藏)组成的网格，算法能够识别并计算出相互连接的油藏区块数量。采用深度优先搜索(DFS)遍历每个区块，并标记已访问过的位置。

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 di erent oil deposits are contained in a grid.
Input
The input le 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 di erent 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.
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题意：
给定一个由“@”和“*”和组成的数组，输出相连的“@”块的个数（八个方向相连）


#include <iostream>
#include <cstring>
using namespace std;
int m,n;
int dx[]={1,1,1,0,0,-1,-1,-1};
int dy[]={-1,0,1,-1,1,-1,0,1};
char grid[110][110];//记录数组
void initial()
{
    memset(grid,' ',sizeof(grid));
    for(int i=0;i<m;i++)
    {
        for(int k=0;k<n;k++)
        {
            cin>>grid[i][k];
        }
    }
}
void dfs(int x,int y)
{
    grid[x][y]='#';//访问过后置为“#”
    int nx,ny;
    for(int i=0;i<8;i++)
    {
        nx=x+dx[i];
        ny=y+dy[i];
        if(nx>=0&&nx<m&&ny>=0&&ny<n&&grid[nx][ny]=='@')
            dfs(nx,ny);
    }
}
int main()
{
    int i,k,sum;
    while(cin>>m>>n)
    {
        sum=0;
        if(m==0&&n==0)
            break;
        initial();
        for(i=0;i<m;i++)
        {
            for(k=0;k<n;k++)
            {
                if(grid[i][k]=='@')
                {
                    dfs(i,k);
                    sum++;
                }
            }
        }
        cout<<sum<<endl;
    }
          return 0;
}