Oil Deposits （深搜）

最新推荐文章于 2025-12-26 16:39:46 发布

原创最新推荐文章于 2025-12-26 16:39:46 发布 · 346 阅读

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文章标签：

#Oil Deposits #深搜 #c++ #acm

本文介绍了一个基于搜索算法的油藏探测问题解决方案。通过构建网格地图并使用深度优先搜索(DFS)来确定不同油藏的数量。文章详细解释了输入输出格式，并提供了一段使用C++实现的DFS代码。

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

题意：这题可以说是搜索中最基础的一题之一。要找出相连在一起的有多少块，
分析：依次枚举，遇到@时就进行搜索，用深搜，广搜应该都行，目的是把相连的@都标记为已访问。
下面给出用DFS的代码，最坑的是最后还有一个换行，要用\n或者%*c。

#include<cstdio>  
#include<cstring>  
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
char map[105][105],mat[105][105],vis[105][105];  
void dfs(int i,int j)  
{  
    if(vis[i][j]||mat[i][j]=='*')  
    return ;  
    vis[i][j]=1;  
    dfs(i-1,j-1);  
    dfs(i-1,j);  
    dfs(i-1,j+1);  
    dfs(i,j-1);  
    dfs(i,j+1);  
    dfs(i+1,j-1);  
    dfs(i+1,j);  
    dfs(i+1,j+1);  
}  
int main()  
{  
    int n,m;  
    while(~scanf("%d %d\n",&n,&m))  
    {  
        memset(map,0,sizeof(map));  
        memset(mat,'*',sizeof(mat));  
        memset(vis,0,sizeof(vis));  
        int i,j,cnt = 0;  
        if(!m&&!n)  
        break;  
        for(i = 0;i<n;i++)  
        {  
            for(j = 0;j<m;j++)  
            {  
                scanf("%c",&map[i][j]);  
                mat[i+1][j+1]=map[i][j];  
            }  
            getchar();  
        }  
        for(i = 1 ;i<=n;i++)  
        {  
            for(j = 1;j<=m;j++)  
            {  
                if(!vis[i][j]&&mat[i][j] == '@')  
                {  
                    dfs(i,j);  
                    cnt++;  
                }  
            }  
        }  
        printf("%d\n",cnt);  
    }  
}