zoj 1709 Oil Deposits

最新推荐文章于 2018-08-01 12:28:35 发布

原创最新推荐文章于 2018-08-01 12:28:35 发布 · 658 阅读

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本文介绍了一种使用深度优先搜索（DFS）算法解决油田勘探问题的方法，通过检测网格中的油藏分布，有效识别油田的存在及其数量。文章详细阐述了输入格式、输出要求以及DFS算法在解决实际问题中的应用。

老报告都突出一个水，这个虽然水但是还算有代表性，就搬运过来好了。

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

简单的DFS题，AC代码如下：

#include<cstdio>
 
char map[101][101];
int m,n;
int dir[8][2]={-1,-1,-1,0,-1,1,0,1,1,1,1,0,1,-1,0,-1};
 
void dfs(int x,int y)
{
     int i,nx,ny;
     map[x][y]='*';
     for(i=0;i<8;i++)
     {
          nx=x+dir[i][0];
          ny=y+dir[i][1];
          if(nx<0 || ny<0 || nx>m || ny>n) continue;
          if(map[nx][ny]=='@')
               dfs(nx,ny);
     }
}
int main()
{
    int i,j;
    int count;
    while(true)
    {
        scanf("%d%d",&m,&n);
        if(m==0) break;
        for(i=0;i<m;i++) scanf("%s",map[i]);
        count=0;
        for(i=0;i<m;i++)
            for(j=0;j<n;j++)
                if(map[i][j]=='@')
                {
                    dfs(i,j);
                    count++;
                }
        printf("%d\n",count);
    }
    return 0;
}