hud1241-DFS求连通块

最新推荐文章于 2025-11-07 10:47:15 发布

原创最新推荐文章于 2025-11-07 10:47:15 发布 · 310 阅读

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本文介绍了一种用于探测地下油藏分布的算法实现。通过深度优先搜索（DFS）方法来确定不同油藏间的连通性，并计算出独立油藏的数量。输入包含一系列网格地图，每个网格代表一个待分析区域，通过字符标记油藏位置。输出则是各区域内独立油藏的总数。

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

0122

搜索入门题，深搜@,深搜次数即为所求联通块个数
代码如下：
#include<iostream>
using namespace std;
char oil[101][101];
int m,n;
void dfs(int x,int y)
{
    oil[x][y]='*';
    int dx,dy;
    int xx,yy;
    for(dx=-1;dx<2;dx++)
    for(dy=-1;dy<2;dy++)
    {
        xx=x+dx;
        yy=y+dy;
        if(xx>=0&&xx<m&&yy>=0&&yy<n&&oil[xx][yy]=='@')
        {
            dfs(xx,yy) ;
        }
    }
}
int main()
{
    int i,j,ans=0;
    while(cin>>m>>n&&m!=0)
    {
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                cin>>oil[i][j];
            }
        }
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                if(oil[i][j]=='@')
                {
                    dfs(i,j);
                    ans++;
                }
            }
        }
        cout<<ans<<endl;
        ans=0;
    }
    return 0;
}

代码注意事项注意边界情况，以及递归调用出口。