C - Oil Deposits （搜索）

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本文介绍了一个用于探测地下油藏分布的算法。通过将目标区域划分为多个正方形地块，并使用传感设备确定每个地块是否含有石油。算法实现了对相邻石油口袋的识别与计数，为地质勘探提供了有效的技术支持。

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C - 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 Input

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

Sample Output

这个题目是直接套用模板，注意标记走过的路就行了；

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int MAX = 1e2 + 10;
int ans,visit[MAX][MAX],m,n;
char mapp[MAX][MAX];

int fx[8]={0,0,-1,1,-1,1,-1,1};
int fy[8]={-1,1,0,0,-1,-1,1,1};


void dfs(int x, int y){
	visit[x][y]=1;//biaoji
	for(int i=0; i<8; i++){
		int xx=x+fx[i],yy=y+fy[i];
		if(xx >= 0 && yy >= 0 && xx < n && yy < m && !visit[xx][yy] && mapp[xx][yy] == '@'){
			dfs(xx,yy);
		}
	}
}

int main(){
	while(cin >> n >> m,(n||m)){
		memset(visit,0,sizeof(visit));
		for(int i=0; i<n; i++){
			cin >> mapp[i];
		}
		ans=0;
		for(int i=0; i<n; i++){
			for(int j=0; j<m; j++){
				if(mapp[i][j]=='@'&&!visit[i][j]){
					ans++;
					dfs(i,j);
				}
			}
		}
		cout << ans << endl;
	}	
	return 0;
}