POJ2965

题目大意:

4*4矩阵,矩阵上每个元素都有两种状态开和关,给定初始状态,和状态转移求从初始状态到目标状态:所有元素都是开状态,的最小步骤数以及步骤.

这题也没多想,位压缩+bfs+路径回溯.

状态数最多2^16个

层数最少:bfs

求步骤:只需Mark每个状态的父状态,回溯一遍即可.

附代码:

#include 
#include 
#include 
#include 
#include 
using namespace std;
//状态转移
const int method[]={0xf888,0xf444,0xf222,0xf111,
              0x8f88,0x4f44,0x2f22,0x1f11,
              0x88f8,0x44f4,0x22f2,0x11f1,
              0x888f,0x444f,0x222f,0x111f};
const int FIN = 0;
const int MAX = 65536;
struct Path
{
	int nStep; //跳转到该状态所需步数。
	int nFatherId; //父结点id
	int nMethod; //当前节点是父节点的第几个孩子
};
Path stPath[MAX];
std::stack Stack;
int InitData()
{
	int i ;
	for (i = 0 ;i < MAX ; i ++)
	{
		stPath[i].nFatherId = -1;
		stPath[i].nMethod = -1;
		stPath[i].nStep = - 1;
	}
	return 0;
}
int input()
{
	int i ,j;
	char ch;
	int nInitState = 0;
	for(i = 0 ;i < 4; i ++)
	{
		for (j = 0 ;j < 4; j ++)
		{
			ch = getchar();
			if('+' == ch)
			{
				nInitState += (int)(pow(2.0,1.0*(15 - i* 4 - j)));
			}
		}
		getchar();
	}
	return nInitState;
}
int bfs(int nInitState)
{
	std::queue Queue;
	//初始状态入队列
	Queue.push(nInitState);
	stPath[nInitState].nStep = 0;
	while(!Queue.empty())
	{
		int nFrontStatue = Queue.front();
		Queue.pop();
		int i ;
		for (i = 0 ;i < 16 ;i ++)
		{
			int nCurStatue = nFrontStatue ^ method[i];

			if (stPath[nCurStatue].nStep == -1)
			{
				Queue.push(nCurStatue);
				stPath[nCurStatue].nStep = stPath[nFrontStatue].nStep + 1;
				stPath[nCurStatue].nMethod = i;
				stPath[nCurStatue].nFatherId = nFrontStatue;
			}
			if (nCurStatue == FIN)
			{
				return nCurStatue;
			}
		}

	}
	return -1;
}
int findPath(int nEnd)
{
	int i;
	i = nEnd;
	Stack.push(stPath[i].nMethod);
	while(1)
	{
		i = stPath[i].nFatherId;
		if (stPath[i].nFatherId == -1)
		{
			break;
		}
		Stack.push(stPath[i].nMethod);
	}
	return 0;
}
int main()
{
	int nInitState = input();
	InitData();
	int nRes = bfs(nInitState);
	findPath(nRes);
	cout<

看到网上有人用递推秒杀此题,感觉自己弱爆了.....

1.一个元素翻转奇数次状态不变,翻转偶数次状态改变

由1容易推到出2

2:要想把第Sij翻转,同时保持第i行和第j列其他元素状态不变,sij本身翻转两次,第i和第j列的其他元素翻转一次.

由把所有状态为关的元素都做一次2操作,记录那些元素状态变化过,变化过的元素个数即最小步骤数.

由以上分析得出,各个元素的步骤顺序其实是无所谓的.把变化的过元素按任意顺序输出即步骤.

附代码:

#include 
#include 
bool mark[4][4];
struct Pos
{
	int x;
	int y;
};
std::queue Ans;
int input()
{
	memset(mark,0,sizeof(mark));
	int i ,j,k;
	char ch;
	for (i = 0 ;i < 4 ; ++ i)
	{
		for (j = 0 ;j < 4 ; ++j)
		{
			ch = getchar();
			if ('+' == ch)
			{
				mark[i][j] = !mark[i][j];
				for(k = 0 ; k < 4 ; k ++)
				{
					mark[i][k] = !mark[i][k];
					mark[k][j] = ! mark[k][j];
				}
			}
		}
		getchar();
	}
	return 0;
}
int main()
{
	int nStep = 0;
	int i,j;
	input();
	for ( i = 0 ; i < 4; ++i)
	{
		for (j = 0 ; j < 4 ; ++j)
		{
			if (mark[i][j])
			{
				nStep ++;
				Pos pos;
				pos.x = i + 1;
				pos.y = j + 1;
				Ans.push(pos);
			}
		}
	}
	std::cout <