Google2015校招在线测试题1----扫雷最少点击次数

标签：c++   算法   

Problem

Minesweeper is a computer game that became popular in the 1980s, and is still included in some versions of the Microsoft Windows operating system. This problem has a similar idea, but it does not assume you have played Minesweeper.

In this problem, you are playing a game on a grid of identical cells. The content of each cell is initially hidden. There are M mines hidden in M different cells of the grid. No other cells contain mines. You may click on any cell to reveal it. If the revealed cell contains a mine, then the game is over, and you lose. Otherwise, the revealed cell will contain a digit between 0 and 8, inclusive, which corresponds to the number of neighboring cells that contain mines. Two cells are neighbors if they share a corner or an edge. Additionally, if the revealed cell contains a 0, then all of the neighbors of the revealed cell are automatically revealed as well, recursively. When all the cells that don‘t contain mines have been revealed, the game ends, and you win.

For example, an initial configuration of the board may look like this (‘*‘ denotes a mine, and ‘c‘ is the first clicked cell):

*..*...**.
....*.....
..c..*....
........*.
..........

There are no mines adjacent to the clicked cell, so when it is revealed, it becomes a 0, and its 8 adjacent cells are revealed as well. This process continues, resulting in the following board:

*..*...**.
1112*.....
00012*....
00001111*.
00000001..

At this point, there are still un-revealed cells that do not contain mines (denoted by ‘.‘ characters), so the player has to click again in order to continue the game.

Input

The first line of the input gives the number of test cases, T. Ttest cases follow. First line of each test case contains one integer N. N lines strings with length N follows containing ‘*‘ and ‘.‘, denotes the Minesweeper initial board.

Output

For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the minimum number of clicks to win.

Limits

1 ≤ T ≤ 100.

Small dataset

1 ≤ N ≤ 50.

Large dataset

1 ≤ N ≤ 300.

思路：简单地说就是给出一个扫雷地图，算出最少点击次数，使得所有非地雷区被访问过。需要一个数组存每个位置周围地雷数目，点击到地雷数为0的时候将自动访问呢周围八个区域，而且会递归访问0的区域。所以解题方法其实就只要是需要深搜，先将地雷数为0的访问，然后深搜访问，到最后再访问那些非地雷非0的区域，就能实现最少点击。

代码：

/*2014-10-16 for google graduates online test
*author:zhuangweibiao
*/
#include<iostream>
#include<fstream>
#include<string>
using namespace std;
string str[302];//存扫雷矩阵
int num[302][302];
bool visit[302][302];
void dfs(int i, int j, int n)
{
	visit[i][j] = true;
	for(int ii = i - 1; ii <= i + 1; ii++)
		for(int jj = j - 1; jj <= j + 1; jj++)
		{
			if(ii >= 0 && ii < n && jj >= 0 && jj < n && str[ii][jj] != '*' && !visit[ii][jj])
			{
				if(num[ii][jj] != 0)
					visit[ii][jj] = true;
				else
					dfs(ii, jj, n);
			}
		}
}
int main()
{
	ifstream ifile("A-large-practice (1).in");
	ofstream ofile("a2.txt");
	int T;
	ifile >> T;
	for(int i = 1; i <= T; i++)
	{
		int n;
		ifile >> n;
		for(int j = 0; j < n; j++)
			ifile >> str[j];
		for(int i = 0; i < n; i++)
			for(int j = 0; j < n; j++)
			{
				if(str[i][j] != '*')
				{
					int count = 0;
					for(int ii = i - 1; ii <= i + 1; ii++)
						for(int jj =j - 1; jj <= j + 1; jj++)
						{
							if(ii >= 0 && ii < n && jj >= 0 && jj < n && str[ii][jj] == '*')
								count++;
						}
						num[i][j] = count;
				}
				else
					num[i][j] = -1;
				visit[i][j] = false;
			}
			int count = 0;
			for(int i = 0; i < n; i++)
				for(int j = 0; j < n; j++)
				{
					if(num[i][j] == 0 && !visit[i][j])
					{
						dfs(i, j, n);
						count++;
					}
				}
			for(int i = 0; i < n; i++)
				for(int j = 0; j < n; j++)
				{
					if(str[i][j] != '*' && !visit[i][j])
						count++;
				}
				ofile << "Case #" << i << ": " << count << endl;
	}
	return 0;
}

Google2015校招在线测试题1----扫雷最少点击次数

标签：c++   算法   

原文地址：http://blog.csdn.net/u012999424/article/details/40145461