pku1562.txt

北京大学 在线acm在线判题系统 一些题目解答

Oil Deposits 
Time Limit:1000MS  Memory Limit:10000K
Total Submit:825 Accepted:510 

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


0
1
2
2


Source
Mid-Central USA 1997

Source

Problem Id:1562  User Id:Drizzle 
Memory:188K  Time:15MS
Language:G++  Result:Accepted

Source 

/*
   农夫三拳@seu
   drizzlecrj.gmail.com
   2007-03-02
*/
#include <iostream>
#include <cstring>

using namespace std;

int m, n;
char map[101][101];
bool visited[101][101];
const int delta[8][2] = 
{
	{-1, 1}, {0, 1},{1, 1},
	{-1, 0}, {1, 0},
	{-1, -1},{0, -1}, {1, -1}
};

bool InRange(int x, int y)
{
	return x >= 0 && x < m && y >= 0 && y < n;
}

void Search(int x, int y)
{
	visited[x][y] = true;
	for(int i = 0; i < 8; i++)
	{
		if(InRange(x + delta[i][0], y + delta[i][1]) && map[x + delta[i][0]][y + delta[i][1]] == '@' && !visited[x + delta[i][0]][y + delta[i][1]])
		{
			Search(x + delta[i][0], y + delta[i][1]);
		}
	}
}

int Solve()
{
	int num = 0;
	for(int i = 0; i < m; i++)
	{
		for(int j = 0; j < n; j++)
		{
			if(map[i][j] == '@' && !visited[i][j])
				Search(i, j), ++num;
		}
	}
	return num;
}

int main()
{
	while(cin >> m >> n && (m || n))
	{
		for(int i = 0; i < m; i++)
		{
			for(int j = 0; j < n; j++)
			{
				cin >> map[i][j];
			}
		}
		memset(visited, false, sizeof(visited));
		cout << Solve() << endl;
	}

	return 0;
}