POJ 2965

The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16008		Accepted: 6035		Special Judge

Description

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

Sample Input

-+--
----
----
-+--

Sample Output

6
1 1
1 3
1 4
4 1
4 3
4 4

AC代码如下（枚举+BFS），不是很快，勉强能够交上：

//1936K	938MS
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
static const int  ROW_COL_NUM = 4;
static const int  GRID_NUM = ROW_COL_NUM * ROW_COL_NUM ;
static const int MAX_STATES = 1 << GRID_NUM ;
int isVisit[MAX_STATES] ; //是否被访问， 0表示没有，否则表示是在经过了几次操作之后访问到的

typedef struct node 
{
	int state; //当前状态
	int pre ;
	int loc ;
	node(int s, int p, int l) : state(s), pre(p), loc(l) {} ;
	node(int s): state(s){pre = -1 ; } ;

}stateNode;


bool isOpened(int state)
{
	if (state == 0) return true ;
	return false ;
}

int switchState(int state, int i)
{
	state ^= (1 << i) ;
	int tmp = i - 4 ;
	while(tmp >= 0)
	{
		state ^= (1 << tmp) ;
		tmp -= 4 ;
	}
	tmp = i + 4 ;
	while (tmp < GRID_NUM)
	{
		state ^= (1 << tmp) ;
		tmp += 4 ;
	}

	tmp = i ;
	while(tmp % ROW_COL_NUM != 0)
	{
		tmp -= 1 ;
		state ^= (1 << tmp) ;
	}
	tmp = i ;
	while(tmp % ROW_COL_NUM != 3)
	{
		tmp += 1 ;
		state ^= (1 << tmp) ;
	}
	return state ;
}

int buildState()
{
	int ret = 0 ;
	char ch ;
	for (int i = 0; i < GRID_NUM ; ++ i)
	{
		cin >> ch ;
		if (ch == '+')
			ret += (1 << i) ;
	}
	return ret ;
}


void bfs(int initState)
{
	memset(isVisit, 0 , sizeof(isVisit)) ;

	vector<stateNode> statesQueue ;
	int qIndex = 0 ;
	statesQueue.push_back(stateNode(initState)) ;

	while(!statesQueue.empty())
	{
		stateNode curState = statesQueue.at(qIndex) ;
		qIndex ++ ;

		for (int i = 0; i < GRID_NUM; ++ i)
		{
			int nextState = switchState(curState.state, i) ;
			if (isVisit[nextState] == 0)
			{
				isVisit[nextState] = isVisit[curState.state] + 1 ;
				stateNode addNode = stateNode(nextState, qIndex - 1, i) ;

				if (isOpened(nextState)) //找到结果
				{
					cout << isVisit[nextState] << endl;
					while(addNode.pre != -1) //打印路径
					{
						cout << addNode.loc/4 + 1<< " " << addNode.loc % 4 + 1<< endl;
						addNode = statesQueue.at(addNode.pre) ;
					}
					return ;
				}
				statesQueue.push_back(addNode) ;
			}
		}
	}
}

int main(int argc, char** argv)
{
	int initValue = buildState() ;
	bfs(initValue) ;
	return 0 ;
}