日志13：bfs例题

最新推荐文章于 2025-07-05 16:03:52 发布

最新推荐文章于 2025-07-05 16:03:52 发布 · 1.1k 阅读

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文章描述了一种技术，GeoSurvComp通过在大区域土地上创建网格并使用感应设备分析每个小区域来检测石油沉积。任务是确定网格中包含的不同石油沉积数量，考虑口袋之间的相邻关系。

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

#include<iostream>
#include<queue>
typedef long long ll;
#define endl '\n'
using namespace std;
int m,n,ans;
char mp[109][109];
int vis[109][109];
int dir[8][2]={1,0,-1,0,0,1,0,-1,1,1,1,-1,-1,1,-1,-1};
struct nm
{
   int x,y;
};nm a;nm b;
queue<nm> q;
int main()
{
   ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);
   while(cin>>n>>m)
   {
       if(n==0) return 0;
       ans=0;
       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=m;j++) cin>>mp[i][j];
       }
       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=m;j++)
           {
               if(mp[i][j]=='@'&&vis[i][j]==0)
               {
                   ans++;
                   vis[i][j]=1;
                   a.x=i,a.y=j;
                   q.push(a);
                   while(!q.empty())
                   {
                       a=q.front();
                       q.pop();
                       for(int k=0;k<=7;k++)
                       {
                           int xx=a.x+dir[k][0];
                           int yy=a.y+dir[k][1];
                           if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&mp[xx][yy]=='@'&&vis[xx][yy]==0)
                           {
                               vis[xx][yy]=1;
                               b.x=xx,b.y=yy;
                               q.push(b);
                           }
                       }
                   }
               }
           }
       }
       cout<<ans<<'\n';
       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=m;j++) vis[i][j]=0;
       }
   }

}