搜索（B - Oil Deposits）

最新推荐文章于 2020-01-27 10:07:07 发布

u013514659

最新推荐文章于 2020-01-27 10:07:07 发布

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本文链接：https://blog.youkuaiyun.com/u013514659/article/details/19995855

本文介绍了一个基于递归深度优先搜索的算法，用于分析矩形土地上的石油分布情况。该算法通过将含有石油的地块标记为已访问来计算独立的石油沉积数量。

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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

题目大意：先输入两个数，代表这片地的长和宽，这片地里有‘@’和‘*’；‘@’代表有石油，‘*’代表无油；求这片地里有几块地方分布着石油。

代码：

#include<string.h>
#include<stdio.h>

using namespace std;
char a[110][110];
int n,m;
void dfs(int x,int y)
{
    if(a[x][y]!='@'||x<0||y<0||x>=n||y>=m)
        return;
    else
    {
        a[x][y]='*';
        dfs(x-1, y-1);
        dfs(x-1, y);
        dfs(x-1, y+1);
        dfs(x, y-1);
        dfs(x, y+1);
        dfs(x+1, y-1);
        dfs(x+1, y);
        dfs(x+1, y+1);
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0 || m==0)
        break;
        else
        {
            for(int i=0;i<n;i++)
            //{
                //for(int j=0;j<m;j++)
                //scanf("%c",&a[i][j]);
                //cin>>a[i][j];
                scanf("%s",a[i]);
            //}
            int count=0;
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<m;j++)
                {
                    if(a[i][j]=='@')
                    {
                        count++;
                        dfs(i,j);

                    }
                }
            }
            printf("%d\n",count);
        }
    }
    return 0;
}