poj 1077 Eight

 

Eight

Time Limit: 1000MS Memory Limit: 65536KTotal 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：5782538bobten20081077Accepted8752K47MSC++7345B2009-08-30 14:52:15如果两个都用，则时间为：5782475bobten20081077Accepted9128K16MSC++7329B2009-08-30 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! - 1using 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ivoid 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;}