HDU5546-Ancient Go

Ancient Go

                                                                           Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
                                                                                                      Total Submission(s): 2292    Accepted Submission(s): 713

Problem Description

Yu Zhou likes to play  Go with Su Lu. From the historical research, we found that there are much difference on the rules between ancient go and modern go.

Here is the rules for ancient go they were playing:

⋅ The game is played on a  8×8  cell board, the chess can be put on the intersection of the board lines, so there are  9×9  different positions to put the chess.
⋅ Yu Zhou always takes the black and Su Lu the white. They put the chess onto the game board alternately.
⋅ The chess of the same color makes connected components(connected by the board lines), for each of the components, if it's not connected with any of the empty cells, this component dies and will be removed from the game board.
⋅ When one of the player makes his move, check the opponent's components first. After removing the dead opponent's components, check with the player's components and remove the dead components.
One day, Yu Zhou was playing ancient go with Su Lu at home. It's Yu Zhou's move now. But they had to go for an emergency military action. Little Qiao looked at the game board and would like to know whether Yu Zhou has a move to kill at least one of Su Lu's chess.

 

Input

The first line of the input gives the number of test cases,  T(1≤T≤100) .  T  test cases follow. Test cases are separated by an empty line. Each test case consist of 9 lines represent the game board. Each line consists of 9 characters. Each character represents a cell on the game board.  ′.′  represents an empty cell.  ′x′  represents a cell with black chess which owned by Yu Zhou.  ′o′  represents a cell with white chess which owned by Su Lu.

 

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  Can kill in one move!!! if Yu Zhou has a move to kill at least one of Su Lu's components.  Can not kill in one move!!! otherwise.

 

Sample Input

   

    2

.......xo
.........
.........
..x......
.xox....x
.o.o...xo
..o......
.....xxxo
....xooo.

......ox.
.......o.
...o.....
..o.o....
...o.....
.........
.......o.
...x.....
........o
   

 

Sample Output

   

    Case #1: Can kill in one move!!!
Case #2: Can not kill in one move!!!

    

     

      Hint
     
In the first test case, Yu Zhou has 4 different ways to kill Su Lu's component.

In the second test case, there is no way to kill Su Lu's component.

    
     
   

 

Source

The 2015 China Collegiate Programming Contest

 

Recommend

wange2014

 

题意：围棋，判断再走一步能不能吃子

解题思路：可以转换为求白棋有多少个气孔 大于等于2个的话就不能被干掉 这样的话我们直接遍历棋盘里面的白棋 然后对他们进行dfs，注意每次搜索时，没子的地方要重新记录

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const LL INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;

char mp[15][15];
int  cnt, flag;
bool vis[15][15];
bool visdot[15][15];
int dir[][2] = { 0,1,1,0,0,-1,-1,0 };

void dfs(int x, int y)
{
	if (x<0 || y<0 || x >= 9 || y >= 9) return;
	if (mp[x][y] == '.')
	{
		if (visdot[x][y]) return;
		visdot[x][y] = 1;
		cnt++;
		return;
	}
	else if (mp[x][y] == 'x') return;
	else
	{
		vis[x][y] = 1;
		for (int i = 0; i<4; i++)
		{
			if (vis[x + dir[i][0]][y + dir[i][1]]) continue;
			dfs(x + dir[i][0], y + dir[i][1]);
		}
	}
}

int main()
{
	int t,cas=0;
	scanf("%d",&t);
	while (t--)
	{
		for (int i = 0; i<9; i++) scanf("%s", mp[i]);
		memset(vis, 0, sizeof(vis));
		flag = 0;
		for (int i = 0; i<9; i++)
		{
			for (int j = 0; j<9; j++)
			{
				if (mp[i][j] == 'o')
				{
					if (vis[i][j]) continue;
					cnt = 0;
					memset(visdot, 0, sizeof(visdot));
					dfs(i, j);
					if (cnt <= 1)
					{
						flag = 1;
						break;
					}
				}
			}
			if (flag)break;
		}
		if (flag) printf("Case #%d: Can kill in one move!!!\n", ++cas);
		else printf("Case #%d: Can not kill in one move!!!\n", ++cas);
	}
	return 0;
}