PKU-1077 Eight (八数码之A*算法)

原题链接 http://poj.org/problem?id=1077

Eight

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

 

A*算法

一般地，节点n的评估函数f(n)可表示如下

　　　　　　　　f(n) = h(n) + d(n)

　　h(n) 是 启发式函数，与启发信息相关。启发函数h(n)的设计非常重要，它无固定的设计模式，主要取决于面向具体问题结构的知识和求解的技巧。d(n)是开始节点到节点n的深度信息。

　　A*算法最为核心的部分，就在评估函数的设计。在A*搜索算法的评估函数中，d(n)表示从起始节点到当前节点的代价，通常用当前节点在搜索树中的深度表示。启发式函数h(n)表示当前节点到目标节点的评估值，是启发式搜索算法最为关键的部分。

　　A*搜索算法的评估函数包括两个约束条件：

　　　　(1) h(n)不大于节点n到目标节点t的实际最小代价h*(n)，即：h(n) <= h*(n)。

　　　　(2) 任意节点的评估值f必须不小于父节点的f值，即f单调递增(如果存在小于的情况，即子结点f值小于父结 点，则从父结点经过子结点再回到父结点，父结点f值会变小……搜索结果会形成一个循环)。

　　满足这两个条件，可以证明A*能够获得问题的最优解。简单证明一下，选择节点x，节点x一定是可扩展节点中的具有最小估计值的节点，如果x是目标节点，则其估计值f(x)就是实际最小代价，不大于其他节点的估计值，又条件（1），所以f(x)不大于其它任意节点的实际最小代价，即为最优解。

一般地，A*搜索算法可以描述如下：A* ( )    open ← {s};    closed ← Ø    while open != Ø do        remove leftmost state from open, call it x        if x is a goal then return success        else            generate children of x            for each child c do                if c is already on open list then                    if c was reached by a shorter path then                        update f(c)               else if c is already on closed list then                    if c was reached by a shorter path then                        remove c from closed list                        add c to open list and update f(c)               else                    assign c a heuristic value f(c)                    add c to open list          put x on closed list and reorder states on open by f    return failure

(注：伪代码中如果扩展结点在closed表中，会比较f值的大小，但是f本身是非递减，从起点开始扩展的点大于等于起点的f值，而后每次又从open去取f最小的值进行扩展，所以 closed中存储的都是当前状态f值最小的点，不会出现所谓的shorter path；等价于如果扩展结点在closed列表，直接continued)

A*算法与BFS：可以这样说，BFS是A*算法的一个特例。对于一个BFS算法，从当前节点扩展出来的每一个节点（如果没有被访问过的话）都要放进队列进行进一步扩展。也就是说BFS的估计函数h永远等于0，没有一点启发式的信息，可以认为BFS是“最烂的”A*算法。而A*搜索算法采用最小代价优先的策略，即在所有将扩展的节点中，选择具有最小代价的节点作为下一个扩展节点。其终止准则是如果选择的节点是目标节点，即停止A*搜索算法。A*()算法中open表是一个优先队列，保存即将要展开的节点。closed表是一个数据结构，例如数组或者链表，保存展开过的节点。s是开始节点。open表可用最小堆实现。(不能在加入open表之前判断是否是目标节点并返回，这样不能保证是由最优路径到达目标节点的)

Source Code

#include <iostream>#include <queue>using namespace std;#define Max 365000   // 9! = 362880struct node {int x, y;char state[10];int cantorValue;int f;    friend bool operator<(node a,node b) {return a.f > b.f;}};char state[Max];  // 0：未查询 1：在open列表中 2：在closed列表中int h[Max];   //估价函数，这里用曼哈顿距离int depth[Max];   //深度int parent[Max], opration[Max];int index;priority_queue<node> Q;node Next, Top;int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1};/*康托展开：X=a[n]*(n-1)!+a[n-1]*(n-2)!+...+a[i]*(i-1)!+...+a[1]*0!例：3 5 7 4 1 2 9 6 8 展开为 98884。因为X=2*8!+3*7!+4*6!+2*5!+0*4!+0*3!+2*2!+0*1!+0*0!=98884.解释： 排列的第一位是3，3以后的序列中比3小的数有2个1，2，以这样的数开始的排列有8!个，因此第一项为2*8!排列的第二位是5，5以后的序列中比5小的数有3个4、1、2，这样的排列有7!个，因此第二项为3*7!  …………以此类推，直至0*0!*/int Cantor(char * buf, int len) {     //bit记录之后的序列中比当前位小的值的个数int cantor[9] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};int i, j, bit;int result = 0;for (i = 0; i < len; i ++) {bit = 0;for (j = i + 1; j < len; j++) {if (buf[i] > buf[j]) {bit ++;}}result += (bit * cantor[len - i - 1]);}return result;}bool isInside(int x, int y) {if (x < 0 || y < 0 || x > 2 || y > 2) {return false;}return true;} //计算曼哈顿距离(只计算1~8，确保每次移动h(n)最多减少1，而g(n)会增加1，这样f = g + h为非递减序列)int goal_state[9][2] = {{0,0}, {0,1}, {0,2}, {1,0}, {1,1}, {1,2}, {2,0}, {2,1}, {2,2}};int manhattan(const char *board){ int i, x, y, num;int dis = 0;for (i = 0; i < 9; i++) {x = i / 3;y = i % 3;num = board[i] - '0';if (num != 9) {dis += (abs(x - goal_state[num - 1][0]) + abs(y - goal_state[num - 1][1]));}}return dis;}//输出路径void print_opration() {int step[50];int len, cur, i;len = cur = 0;while (parent[cur] != -1) {step[len ++] = opration[cur];cur = parent[cur];}for (i = len - 1; i >= 0; i--) {switch (step[i]){case 0: printf("d");break;case 1:printf("u");break;case 2:printf("r");break;case 3:printf("l");break;}}printf("\n");}bool A_star(node start) {int i, j, nextX, nextY;int a, b;char temp;while (!Q.empty()) {Q.pop();}Q.push(start);while (!Q.empty()) {Top = Q.top();Q.pop();if (Top.cantorValue == 0) {print_opration();return true;}for (i = 0; i < 4; i++) {nextX = Top.x + dir[i][0];nextY = Top.y + dir[i][1];if (!isInside(nextX, nextY)) {continue;}            Next.x = nextX;Next.y = nextY; for (j = 0; j < 9; j ++) {Next.state[j] = Top.state[j];}a = Next.x * 3 + Next.y;b = Top.x * 3 + Top.y;temp = Next.state[a];Next.state[a] =  Next.state[b];Next.state[b] = temp;Next.cantorValue = Cantor(Next.state, 9);if (state[Next.cantorValue] == 0) {                //不在open也在不closed中，插入openparent[Next.cantorValue] = Top.cantorValue;opration[Next.cantorValue] = i;h[Next.cantorValue] = manhattan(Next.state);depth[Next.cantorValue] = depth[Top.cantorValue] + 1;Next.f = depth[Next.cantorValue] + h[Next.cantorValue];Q.push(Next);state[Next.cantorValue] = 1;} //在open列表中(说明h[]即曼哈顿距离肯定已经计算过)，并且新的f值小于原先的f, 因为f = g + h,而h不变，所以只需要比较g的大小else if (state[Next.cantorValue] == 1 && (depth[Top.cantorValue] + 1 < depth[Next.cantorValue])) {   parent[Next.cantorValue] = Top.cantorValue;opration[Next.cantorValue] = i;depth[Next.cantorValue] = depth[Top.cantorValue] + 1;Next.f = depth[Next.cantorValue] + h[Next.cantorValue];Q.push(Next);                 //直接插入新值, 有冗余，但不会错}//在closed列表中，比较两个X的估价值，注意是同一个节点的两个不同路径的估价值else if (state[Next.cantorValue] == 2 && (depth[Top.cantorValue] + 1 < depth[Next.cantorValue])) {/*parent[Next.cantorValue] = Top.cantorValue;opration[Next.cantorValue] = i;depth[Next.cantorValue] = depth[Top.cantorValue] + 1;Next.f = depth[Next.cantorValue] + h[Next.cantorValue];Q.push(Next);state[Next.cantorValue] = 1;*/continue;}}state[Top.cantorValue] = 2;}return false;}int main() {int i;char init[10];index = 0;memset(h, -1, sizeof(h));memset(state, 0, sizeof(state));node start;for (i = 0; i < 9; i++) {cin >> init[i];if (init[i] == 'x') {init[i] = '0' + 9;start.x = i / 3;start.y = i % 3;}}for (i = 0; i < 9; i++) {start.state[i] = init[i];}start.cantorValue = Cantor(init, 9);    state[start.cantorValue] = 1;depth[start.cantorValue] = 0;h[start.cantorValue] = manhattan(start.state);start.f = h[start.cantorValue] + depth[start.cantorValue];parent[start.cantorValue] = -1;opration[start.cantorValue] = -1;if (!A_star(start)) {printf("unsolvable\n");}return 0;}