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最新推荐文章于 2022-09-08 19:54:10 发布

Dwylkz

最新推荐文章于 2022-09-08 19:54:10 发布

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分类专栏： Breadth-First-Search Simulation Pretreatement 文章标签： c function ini c++

本文链接：https://blog.youkuaiyun.com/dwylkz/article/details/7918048

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本文探讨了如何通过预处理所有可能的状态，并计算从初始状态到任意状态所需的最小步数，适用于九宫格状态转换问题。

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/*
   As it's only 9^9 status at most(in fact, it's not that much),
   We can calculate all the status,
   which can be reached from initial status 123456789 and
   the number of steps not larger than 9, in advance.
*/
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <cctype>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
//STL-Alias
#define UN(c, a) unique(c, a)
#define MS(c, a) memset(c, a, sizeof c)
#define FLC(c, a ,b) fill(c, c + a, b)
#define LOS(c, a, b) lower_bound(c, a, b)
#define UPS(c, a, b) upper_bound(c, a, b)
//Syntax-Alias
#define Rep(c, a, b) for (int c = (a); c < (b); c++)
#define Nre(c, a, b) for (int c = (a); c > (b); c--)
//B-OP
#define LM(x, y) ((x) << (y))
#define RM(x, y) ((x) >> (y))
#define GB(x, y) (RM(x, y) & 1)
#define SB(x, y) (LM(1, y) | x)
#define SBI(x, a, b) ((LM(1, b) - LM(1, a)) | x)
#define CBI(x, a, b) (~(LM(1, b) - LM(1, a)) & x)
//DEBUG
#define FK puts("Fu*k here!")
#define PA(s, c, a, b, p, f){\
	printf(s);\
	Rep(c, a, b) printf(p, (f));\
	puts("");}
//Constant
#define MOD (1000000000LL)
#define MAXN (1 << 16)
//Type-Alias
typedef long long LL;
typedef long double LD;
typedef int AI[MAXN];
typedef double AD[MAXN];
typedef LL ALL[MAXN];
typedef LD ALD[MAXN];


//H[s] as Minimum steps to state s 
map<LL, int> H; 
//Four kinds of rotate
int mr[4][4] = {{0, 1, 4, 3}, {1, 2, 5, 4}, {3, 4, 7, 6}, {4, 5, 8, 7}}; 
//Encode
int EC(int b[3][3])
{
	int s = 0;
	Rep(i, 0, 3) Rep(j, 0, 3)
		s = s * 10 + b[i][j];
	return s;
}
//Decode
void DC(int s, int b[3][3])
{
	Nre(i, 2, -1) Nre(j, 2, -1)
	{
		b[i][j] = s % 10;
		s /= 10;
	}
}
//Rotate function, d indicate direction
int rot(LL s, int *m, int d)
{
	int cp = 0, x = m[cp] / 3, y = m[cp] % 3, b[3][3];
	DC(s, b);
	int t = b[x][y];
	Rep(j, 1, 4)
	{
		int np = (cp + d + 4) % 4, tx = m[np] / 3, ty = m[np] % 3;
		b[x][y] = b[tx][ty];
		cp = np;
		x = tx;
		y = ty;
	}
	b[x][y] = t;
	return EC(b);
}
//Pretreatment 
void H_ini()
{
	H.clear();
	queue<LL> Q;
	LL u = 123456789;
	H[u] = 1;
	Q.push(u);
	while (!Q.empty())
	{
		u = Q.front(); Q.pop();
		if (H[u] > 8) continue;
		Rep(i, 0, 4) Rep(j, -1, 2)
		{
			LL v = rot(u, mr[i], j);
			if (!H[v])
			{
				H[v] = H[u] + 1;
				Q.push(v);
			}
			j++;
		}
	}
}


LL s;


int main()
{
#if 1
	freopen("test.txt", "r", stdin);
#endif
	//Initialize
	int Cas = 1;
	H_ini();
	//Solve
	while (scanf("%lld", &s), s)
	{
		printf("%d. ", Cas++);
		LL p = s / MOD;
		s %= MOD;
		if (H[s] && H[s] - 1 <= p) printf("%d\n", H[s] - 1);
		else puts("-1");
	}
	return 0;
}