L - Oil Deposits-优快云博客

这篇博客讨论了一种经典的问题——八连通问题，该问题在油田探测中应用。给定一个矩形区域的网格，其中*表示没有油，@表示油井，任务是找出所有独立的油田。每个油田中的油井彼此相邻，可以是水平、垂直或对角线方向。博主提供了一个C++解决方案，通过深度优先搜索遍历网格，统计油田数量。算法在遇到油井时将其标记并进行深度搜索，最终输出油田的总数。

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L - Oil Deposits

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

Inputcopy	Outputcopy
1 1 * 3 5 @@* @ @@* 1 8 @@***@ 5 5 ***@ @@@ @*@ @@@@ @@**@ 0 0	0 1 2 2

Inputcopy

Outputcopy

1 1 * 3 5 *@*@* **@** *@*@* 1 8 @@****@* 5 5 ****@ *@@*@ *@**@ @@@*@ @@**@ 0 0

0 1 2 2

Sponsor

经典的八连通问题，对于这个题，我们遍历整个图，遇到@变成*，遍历所有能够达到的位置，最后只要看进行了多少次深搜，就有多少个油田。

/*Where there is light, in my heart.*/
/*SUMMER_TRAINING DAY 24*/
#include <bits/stdc++.h>
#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
//
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//#define INF 0x3f3f3f
#define ll long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define unmap(a,b) unordered_map<a,b>
#define unset(a) unordered_set<a>
#define F first
#define S second·
#define pb push_back
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define _rep(i, a, b) for (int i = (a); i >= (b); --i)
#define mode 1e4+7
#define pi acos(-1)
typedef double db;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef vector<int> vi;
const int N = 2e5+5;
//
int n,m;
char mp[105][105];
void dfs(int x,int y){
    mp[x][y]='*';
    for(int dx=-1;dx<=1;dx++)
    for(int dy=-1;dy<=1;dy++){
        int nx=x+dx;
        int ny=y+dy;
        if(nx>=0 && nx<m && ny>=0 && ny<n && mp[nx][ny]=='@')
        dfs(nx,ny);
    }
}
signed main(){
    int ans;
    while(cin>>m>>n && m!=0){
        ans=0;
        for(int i=0;i<m;i++){
            cin>>mp[i];
        }
        for(int i=0;i<m;i++)
        for(int j=0;j<n;j++){
            if(mp[i][j]=='@'){
                dfs(i,j);
                ans++;
            }
        }
        cout<<ans<<endl;
    }
}
//made by shun 20220727