深搜1 poj 1562

Oil Deposits

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9628 Accepted: 5275

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 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

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

每次8个方向搜索出与当前节点邻近的节点，直到把与这个节点在同一区域内的所有节点全都标记上，

在main里遍历有多少个不同的区域。。。。ans即为答案。。。

这道题是dfs和  bfs  都能

AC的大水题。。。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=110;char map[maxn][maxn];bool vis[maxn][maxn];int n,m;int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};void dfs(int x,int y){vis[x][y]=1;for(int i=0;i<8;i++){int nx=x+dir[i][0];int ny=y+dir[i][1];if(nx<1||nx>n) continue;if(ny<1||ny>m) continue;if(vis[nx][ny]) continue;if(map[nx][ny]=='*') continue;dfs(nx,ny);}}int main(){while(cin>>n>>m,n&&m){for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)cin>>map[i][j];memset(vis,0,sizeof(vis));int ans=0;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)if(map[i][j]=='@'&&!vis[i][j]){ans++;dfs(i,j);}cout<<ans<<endl;}return 0;}