poj 1077 Eight

Eight

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9915		Accepted: 4303		Special Judge

Description
The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one
tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:

1  2  3  4

5  6  7  8

9 10 11 12

13 14 15  x

where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:

1  2  3  4    1  2  3  4    1  2  3  4    1  2  3  4

5  6  7  8    5  6  7  8    5  6  7  8    5  6  7  8

9  x 10 12    9 10  x 12    9 10 11 12    9 10 11 12

13 14 11 15   13 14 11 15   13 14  x 15   13 14 15  x

r->           d->           r->

The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.

Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and

frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).

In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three

arrangement.
Input
You will receive a description of a configuration of the 8 puzzle. The description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within
a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle

1  2  3

x  4  6

7  5  8

is described by this list:

1 2 3 x 4 6 7 5 8

Output
You will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The
string should include no spaces and start at the beginning of the line.
Sample Input

2  3  4  1  5  x  7  6  8

Sample Output

ullddrurdllurdruldr

Source
South Central USA 1998

/*好久没有遇到这么大快人心的题了，虽然做了一天多，但是这次一改以往遇到陌生题比较浮躁的态度，认真去学习了相关知识。
通过这道题学习了A*算法，以及排列数一种常见的hash算法，并实际使用和操作了最大堆，可以说是收获颇多
*/
/*注意程序中的两个优化：
如果两个优化都没有，则会TLE
如果只用优化2，而不用优化1，所用时间为，同样会TLE
如果只用优化1，而不用优化2，所用时间为：
5782538	bobten2008	1077	Accepted	8752K	47MS	C++	7345B	2009-08-30 14:52:15
如果两个都用，则时间为：
5782475	bobten2008	1077	Accepted	9128K	16MS	C++	7329B	2009-08-30 14:41:28
所以可见优化二的作用是决定性的
*/
#include <iostream>
#define MAX_N 362880 //利用阶乘逆序数来计算Hash值，最大hash值是9! - 1
using namespace std;
//0-9的阶乘，用来计算哈希值
int fac[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
int temp[10];
//记录1 - 9 九个位置可以移动的方向的个数
int dirN[10] = {0, 2, 3, 2, 3, 4, 3, 2, 3, 2};
//记录1 - 9 九个位置可以移动的方向的种类
char dirT[10][6] =  {
{' '},
{' ', 'r', 'd'},
{' ', 'l', 'r', 'd'},
{' ', 'l', 'd'},
{' ', 'u', 'r', 'd'},
{' ', 'u', 'l', 'r', 'd'},
{' ', 'u', 'l', 'd'},
{' ', 'u', 'r'},
{' ', 'u', 'l', 'r'},
{' ', 'u', 'l'}};
int heap[MAX_N + 5];
int heapSize = 0;
int initState, finalState = 123456789, curPos;
struct node
{
int type;   //0 尚未访问, 1 已经在待考察队列中 2 被访问过并从待访问队列中删除了
int heapPos; //当前状态在堆中的位置
int state;  //当前状态，如123456789
char dType; //上一个状态到当前状态所走的方向
int diff;   //A*算法中的h()用当前状态和终止状态之间的different * 10 表示（不计空格）
int step;   //A*算法中的g(),表示当前遍历的深度
int fVal;   //diff和step的和，堆算法主要基于这个指标
node()
{
step = 0;
type = 0;
}
}hashT[MAX_N + 5]; //哈希表
int eP(int l)  //计算10的l次方
{
int res = 1;
for(int i = 1; i <= l; i++)
res *= 10;
return res;
}
int getNum(int which, int num) //通过当前状态取3 × 3方格中的第which个数
{
if(which > num % 10)
return num / eP(9 - which + 1) % 10;
else
return num / eP(9 - which) % 10;
}
//以下均是移动空格的位置
int moveL(int state) //由当前状态左移得到新的状态
{
return state - 1;
}
int moveR(int state) //由当前状态右移得到新的状态
{
return state + 1;
}
int moveU(int state) //由当前状态下移得到新的状态
{
int t0 = 9 - state % 10 + 1;
int t1 = state / eP(t0);
int t2 = t1 % 1000;
t1 = t1 - t2 + (t2 % 100) * 10 + t2 / 100;
t1 = t1 * eP(t0);
return t1 + state % eP(t0) - 3;
}
int moveD(int state) //由当前状态上移得到新的状态
{
int t0 = 7 - state % 10;
int t1 = state / eP(t0);
int t2 = t1 % 1000;
t1 = t1 - t2 + (t2 % 10) * 100 + t2 / 10;
t1 = t1 * eP(t0);
return t1 + state % eP(t0) + 3;
}
int getRevNum(int state) //得到当前状态的逆序数，最后一位表示空格，不计
{
int blank = state % 10;
int total = 0;
int p = 0;
for(int id = 1; id <= 9; id++)
{
if(id == blank)
continue;
int n = getNum(id, state);
temp[p++] = n;
for(int k = p - 2; k >= 0; k--)
if(temp[k] >= n)
total++;
}
return total;
}
//通过当前状态得到其对应的哈希值
//这里计算哈希值采用1-9每位的逆序数乘以（位数减一的阶乘）并取和的方法，这样
//可以保证最大的哈希值为9! - 1且没个状态的哈希值都唯一
int getHashNum(int state)
{
int blank = state % 10;
int hashVal = 0;
int p = 0, total;
int pp = 0; //pp记录阶乘的位置
for(int id = 1; id <= 9; id++)
{
pp++;
if(id == blank)
continue;
int n = getNum(id, state);
temp[p++] = n;
total = 0;
for(int k = p - 2; k >= 0; k--)
{
if(temp[k] >= n)
total++;
}
hashVal += total * fac[p - 1];
}
// 空格作为最小值计算
return hashVal + (blank - 1) * fac[blank - 1];

}
//计算两个状态的距离
int getDiff(int state1, int state2)
{
int blank1 = state1 % 10, blank2 = state2 % 10, i, diff = 0;
if(blank1 != blank2) diff += 2;
for(i = 1; i <= 9; i++)
{
if(i != blank1 && i != blank2 && getNum(i, state1) != getNum(i, state2))
diff++;
}
return diff;
}
void swapHeapV(int i, int j)
{
int temp = heap[i];
heap[i] = heap[j];
heap[j] = temp;
hashT[heap[i]].heapPos = i;
hashT[heap[j]].heapPos = j;
}
//从i结点保证堆的性质
void maxHeapify(int i)
{
int l = 2 * i, r = 2 * i + 1, largest;
if(l <= heapSize && hashT[heap[l]].fVal < hashT[heap[i]].fVal)
largest = l;
else
largest = i;
if(r <= heapSize && hashT[heap[r]].fVal < hashT[heap[largest]].fVal)
largest = r;

if(largest != i)
{
swapHeapV(i, largest);
maxHeapify(largest);
}
}
//取堆中的最大值
int getHeapMax()
{
return heap[1];
}
//取堆中的最大值，并删除这个值，需要用maxHeapify(1)来保持最大堆的性质
int extracMaxHeap()
{
if(heapSize < 1)
return -1;
int maxv = heap[1];
heap[1] = heap[heapSize];
hashT[heap[1]].heapPos = 1;
heapSize--;
maxHeapify(1);
return maxv;
}
//向对中插入结点i
void heapInsert(int i)
{
heapSize++;
heap[heapSize] = i;
hashT[i].heapPos = heapSize;
int pos = heapSize;
while(pos > 1 && hashT[heap[int(pos / 2)]].fVal > hashT[heap[pos]].fVal)
{
swapHeapV(pos, int(pos / 2));
pos = pos / 2;
}
}
//当结点i的权值发生变化时，需要更新堆的结构以保证最大堆的性质
void upMoveHeap(int i)
{
int pos = i;
while(pos > 1 && hashT[heap[int(pos / 2)]].fVal > hashT[heap[pos]].fVal)
{
swapHeapV(pos, int(pos / 2));
pos = pos / 2;
}
}
//递归打印结果
void printRes(int curState)
{
int hashVal = getHashNum(curState);
char dType = hashT[hashVal].dType;
if(dType != 's')
{

if(dType == 'l')
{
printRes(moveR(curState));
cout<<'l';
}
else if(dType == 'r')
{
printRes(moveL(curState));
cout<<'r';
}
else if(dType == 'u')
{
printRes(moveD(curState));
cout<<'u';
}
else if(dType == 'd')
{
printRes(moveU(curState));
cout<<'d';
}
}
}
//计算的核心控制函数
void getBest()
{
//计算并初始化初始结点的相关性质
int h = getHashNum(initState), d, curHashVal, curState, xPos, nextState, nextHashVal, nextHeapPos;
char type;
hashT[h].state = initState;
hashT[h].type = 1;
hashT[h].dType = 's';
hashT[h].step = 0;
//将初始节点插入堆中
heapInsert(h);
while(true)
{
//取当前堆中最大的结点
curHashVal = extracMaxHeap();
//设置当前状态为从待考察队列中删除
hashT[curHashVal].type = 2;

curState = hashT[curHashVal].state;

//到达终止结点
if(curState == finalState)
{
printRes(curState);
cout<<endl;
break;
}
xPos = curState % 10;

//遍历空格可以走的方向
for(d = 1; d <= dirN[xPos]; d++)
{
type = dirT[xPos][d];
if(type == 'r')
{
//从当前节点向右走到达新的状态
nextState = moveR(curState);
//计算新状态的哈希值
nextHashVal = getHashNum(nextState);
nextHeapPos = hashT[nextHashVal].heapPos;
//如果此状态已经被访问过
if(hashT[nextHashVal].type != 0)
{
//优化1：更新此状态的g函数，即遍历的深度
if(hashT[curHashVal].step + 1 < hashT[nextHashVal].step)
{
hashT[nextHashVal].step = hashT[curHashVal].step + 1;
hashT[nextHashVal].fVal = hashT[nextHashVal].step + hashT[nextHashVal].diff;
//更新堆的结构以保证最大堆的性质
upMoveHeap(nextHeapPos);
}
continue;
}
//记录方向
hashT[nextHashVal].dType = 'r';
}
else if(type == 'l')
{
nextState = moveL(curState);
nextHashVal = getHashNum(nextState);
nextHeapPos = hashT[nextHashVal].heapPos;
if(hashT[nextHashVal].type != 0)
{
if(hashT[curHashVal].step + 1 < hashT[nextHashVal].step)
{
hashT[nextHashVal].step = hashT[curHashVal].step + 1;
hashT[nextHashVal].fVal = hashT[nextHashVal].step + hashT[nextHashVal].diff;
upMoveHeap(nextHeapPos);
}
continue;
}
hashT[nextHashVal].dType = 'l';
}
else if(type == 'u')
{
nextState = moveU(curState);
nextHashVal = getHashNum(nextState);
nextHeapPos = hashT[nextHashVal].heapPos;
if(hashT[nextHashVal].type != 0)
{
if(hashT[curHashVal].step + 1 < hashT[nextHashVal].step)
{
hashT[nextHashVal].step = hashT[curHashVal].step + 1;
hashT[nextHashVal].fVal = hashT[nextHashVal].step + hashT[nextHashVal].diff;
upMoveHeap(nextHeapPos);
}
continue;
}
hashT[nextHashVal].dType = 'u';
}
else if(type == 'd')
{
nextState = moveD(curState);
nextHashVal = getHashNum(nextState);
nextHeapPos = hashT[nextHashVal].heapPos;
if(hashT[nextHashVal].type != 0)
{
if(hashT[curHashVal].step + 1 < hashT[nextHashVal].step)
{
hashT[nextHashVal].step = hashT[curHashVal].step + 1;
hashT[nextHashVal].fVal = hashT[nextHashVal].step + hashT[nextHashVal].diff;
upMoveHeap(nextHeapPos);
}
continue;
}
hashT[nextHashVal].dType = 'd';
}

//存储并计算新状态的相关信息
hashT[nextHashVal].state = nextState;
//g函数：遍历的深度加一
hashT[nextHashVal].step = hashT[curHashVal].step + 1;
//h函数：优化2，将h函数乘以10可以降低深度的影响，从而提高A*算法的效率，但是结果一般不是最优的了
hashT[nextHashVal].diff = getDiff(nextState, finalState) * 10;
//计算：f函数即g函数和h函数之和
hashT[nextHashVal].fVal = hashT[nextHashVal].diff + hashT[nextHashVal].step;
//标记此状态已经加入待考察队列
hashT[nextHashVal].type = 1;
//将新状态加入堆（待考察队列）
heapInsert(nextHashVal);
}
}
}
int main()
{

char ch;
int i;
initState = 0;
for(i = 1; i <= 9; i++)
{
cin>>ch;
if(ch == 'x')
curPos = i;
else
initState = initState * 10 + int(ch - '0');
}
initState = initState * 10 + curPos;
int revInit = getRevNum(initState);
if(revInit % 2 != 0)
cout<<"unsolvable"<<endl;
else if(initState == finalState)
cout<<"lr"<<endl;
else
getBest();

return 0;
}