八数码问题高效算法-HDU 1043

ACM算法日常

发布于 2018-08-07 19:39:58

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八数码问题是bfs中的经典问题，经常也会遇到与其相似的题目。用到的思想是bfs+hash；主要是由于状态分散，无法直接用一个确定的数表示。所以导致bfs时，无法去判断一个状态是否已经被搜过。也无法用d数组去求解。这个时候就需要用到hash的方法判断当前状态是否已经被搜过。并按照搜索的顺序给每个状态编号（用这个编号代替对应的状态，与状态一一对应，为了求d[]），将所有的状态存起来，供hash查找。

值得注意的是，八数码问题的状态正好是所有的全排列（9!），由于这个特殊的原因，可以直接用每个状态对应的是第几个排列来给状态编号。这样的一一对应，可以做到当给定一个状态时，就能过直接计算出这个状态对应的编号，这样就可以直接用一个数组标记这个状态是否被搜索过，就不需要利用哈希技术来查找这个状态是否已经被搜过。

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1043

Problem 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, several descriptions of configuration of the 8 puzzle. One 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. Do not print a blank line between cases.

Sample Input

2  3  4  1  5  x  7  6  8

Sample Output

ullddrurdllurdruldr

代码G++（需要使用queue队列结构）：耗时68ms

说明：本解法最精彩的地方在于hash值的计算，整体思路非常简单，倒推算出所有的hash。

#include <stdio.h>
#include <iostream>
#include <queue>
using namespace std;

//简要记录步信息
typedef struct Step_s
{
    //上一步到这一步的方向
    char dir;
    //上一步的hash值
    int parent;
} Step;

//倒推步法时记录的队列节点
typedef struct Node_s
{
    //魔板信息
    int board[9];
    //x所在位置
    int x_index;
    //下一步的hash值
    int child;
} Node;

//四个方向
int dir[4][2] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };

//用于hash值计算
int fac[10];

//所有步法集合
Step step_set[370000];

//设置阶乘值，用于后面的hash值计算
void set_fac()
{
    fac[0] = 1;
    for (int i = 1; i <= 8; ++i) {
        fac[i] = fac[i - 1] * i;
    }
}

//对于任意的board，算出对应的hash值
int board_hash(int board[])
{
    int i, j, ans = 0, k;
    for (i = 0; i < 9; ++i) {
        k = 0;
        for (j = i + 1; j < 9; j++) {
            if (board[i] > board[j]) {
                k++;
            }
        }
        ans += k * fac[8 - i];
    }
    return ans;
}

//广度优先搜索
void bfs(int finish[])
{
    queue<Node> Q;
    Node current;
    int tx, ty, temp, t = 0;
    //记录这个节点的魔板信息
    for (int i = 0; i < 9; ++i) {
        current.board[i] = finish[i];
    }
    //x在current.board[8]的位置上,child和parent的0表示结束和存在路径
    current.x_index = 8; current.child = 0;
    step_set[current.child].parent = 0;

    Q.push(current);
    while (!Q.empty())
    {
        //取得队列的第一个元素
        current = Q.front(); Q.pop();
        //从4个方向开始前进
        for (int i = 0; i < 4; ++i) {
            Node next = current;
            //取得该方向的下一个位置
            tx = current.x_index % 3 + dir[i][0];
            ty = current.x_index / 3 + dir[i][1];
            //防止出界
            if (tx >= 0 && ty >= 0 && tx < 3 && ty < 3) {
                next.x_index = ty * 3 + tx;
                //交换x的位置
                temp = next.board[next.x_index];
                next.board[next.x_index] = next.board[current.x_index];
                next.board[current.x_index] = temp;
                //计算hash值
                next.child = board_hash(next.board);
                //查看是否重复
                if (step_set[next.child].parent == -1) {
                    //上一步的hash值
                    step_set[next.child].parent = current.child;
                    //设置方向
                    if (i == 0)step_set[next.child].dir = 'l';
                    if (i == 1)step_set[next.child].dir = 'r';
                    if (i == 2)step_set[next.child].dir = 'u';
                    if (i == 3)step_set[next.child].dir = 'd';
                    //每次前进则放进队列（广度优先搜索）
                    Q.push(next);
                }
            }
        }
    }
}

int main()
{
    int i, j, s, board_input[10], finish[9];
    char ch[50];

    /*完成状态是[1,2,3,4,5,6,7,8,9]*/
    for (i = 0; i < 9; ++i) {
        finish[i] = i + 1;
    }
    //初始化步数集合的父hash值
    for (i = 0; i < 370000; ++i) {
        step_set[i].parent = -1;
    }
    //设置阶乘函数，用于后面的hash值计算
    set_fac();
    //开始倒推，设置所有的步法
    bfs(finish);
    //输入魔板，gets函数能够输入整行
    while (gets(ch)) {
        for (i = 0, j = 0; ch[i] != '\0'; i++) {
            if (ch[i] == 'x') {
                board_input[j++] = 9;
            }
            else if (ch[i] >= '0' && ch[i] <= '8') {
                board_input[j++] = ch[i] - '0';
            }
        }
        //计算输入的魔板hash值
        s = board_hash(board_input);
        //如果不存在到达这种魔板的路线
        if (step_set[s].parent == -1) {
            printf("unsolvable\n");
            continue;
        }
        //输出所有的步法
        while (s != 0) {
            printf("%c", step_set[s].dir);
            s = step_set[s].parent;
        }
        printf("\n");
    }
}

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原始发表：2017-12-22，如有侵权请联系 cloudcommunity@tencent.com 删除

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