HDU - 1241 Oil Deposits

最新推荐文章于 2021-01-24 13:36:02 发布

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本文介绍了一个简单的算法，用于计算二维网格中表示油藏的连通区域数量。通过深度优先搜索（DFS）遍历网格，将相邻的油藏区块标记为已访问，并计数独立的油藏数量。

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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
这道题比较简单，求@有几个联通块。直接搜索第一个@，然后dfs周围，找到@标记为*表示访问过。

#include <iostream>

using namespace std;

char Map[105][105];
int dd[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}};
int m,n,sum;

void dfs(int x,int y)
{
    for(int i=0;i<8;i++)
    {
        int xx=x+dd[i][0];
        int yy=y+dd[i][1];
        if(xx>=1&&xx<=m&&yy>=1&&yy<=n)
        {
            if(Map[xx][yy]=='@')
            {
                Map[xx][yy]='*';
                dfs(xx,yy);
            }
        }
    }
}
int main()
{
    while(cin>>m>>n)
    {
        if(m==0&&n==0)break;
        sum=0;
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                cin>>Map[i][j];
            }
        }
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(Map[i][j]=='@')
                {
                    sum++;
                    Map[i][j]='*';
                    dfs(i,j);
                }
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}