Oil Deposits

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16655		Accepted: 8917

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.

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.

are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

1 1

*

3 5

*@*@*

**@**

*@*@*

1 8

@@****@*

5 5

****@

*@@*@

*@**@

@@@*@

@@**@

0 0

0

1

2

2

本题是经典的DFS题，利用DFS求连通块的数量，DFS利用的是递归实现

#include<iostream>

using namespace std;

const int maxn = 100+5;

char pic[maxn][maxn];   //图表

int m, n, idx[maxn][maxn];   //连通分量编号矩阵

//用DFS的方法为连通分量矩阵编号

void DFS(int r, int c, int id)

{

	if(r<0||r>m||c<0||c>=n)return; //overload

	if(idx[r][c]>0||pic[r][c]!='@')return; //no '@' or 已经访问

	idx[r][c]=id; //连通分量编号

	for(int dr=-1; dr<=1;dr++)

		for(int dc=-1;dc<=1;dc++)

			if(dr!=0||dc!=0)DFS(r+dr, c+dc, id);

}

int main()

{

	while(cin>>m>>n&&m!=0)

	{

		for(int i=0;i<m;i++)

			for(int j=0;j<n;j++)

				cin>>pic[i][j];

		for(int i=0;i<m;i++)

			for(int j=0;j<n;j++)

				idx[i][j]=0;

		int cnt=0;

		for(int i=0;i<m;i++)

			for(int j=0;j<n;j++)

				if(idx[i][j]==0&&pic[i][j]=='@')DFS(i,j,++cnt);

		cout<<cnt<<endl;

	}

	return 0;

}