NC51032 八数码

题目链接

题目

题目描述

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.

输入描述

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

输出描述

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.

示例1

输入

 2  3  4  1  5  x  7  6  8 

输出

ullddrurdllurdruldr

题解

知识点：BFS。

很显然用bfs搜索，但状态保存是个问题，可以用c++自带的map进行状态保存，也可以用康托展开对局面字符串转化为整型保存（我居然还不会qwq）。

要注意的是，数据再复杂点可以卡普通的bfs，这时候需要优化搜索，可以用双向bfs或者A*，可以节省大量时间。这里我用了双向bfs（因为不会A*2333）。

最后注意无解情况可能超时，建议计数跳出（也有逆序对的方法）。

时间复杂度 \(O(?)\)

空间复杂度 \(O(1)\)

代码

#include <bits/stdc++.h>

using namespace std;

struct node {
    string s;
    int x, y;
};
const int dir[4][2] = { {-1,0},{0,-1},{0,1},{1,0} };
char dirs[4] = { 'u','l','r','d' };

string bfs(node init, node ans) {

    map<string, string> vis1, vis2;
    vis1[init.s] = "";
    vis2[ans.s] = "";

    queue<node> q1, q2;
    q1.push(init);
    q2.push(ans);

    int cnt = 0;
    while (!q1.empty() && !q2.empty()) {
        if (cnt >= 10000) break;
        cnt++;
        node a = q1.front();
        node b = q2.front();
        q1.pop();
        q2.pop();
        if (vis2.count(a.s)) return vis1[a.s] + vis2[a.s];
        else if (vis1.count(b.s)) return vis1[b.s] + vis2[b.s];
        for (int i = 0;i < 4;i++) {
            node aa;
            aa.x = a.x + dir[i][0];
            aa.y = a.y + dir[i][1];
            if (aa.x >= 0 && aa.x < 3 && aa.y >= 0 && aa.y < 3) {
                aa.s = a.s;
                swap(aa.s[a.x * 3 + a.y], aa.s[aa.x * 3 + aa.y]);
                if (!vis1.count(aa.s)) vis1[aa.s] = vis1[a.s] + dirs[i], q1.push(aa);
            }

            node bb;
            bb.x = b.x + dir[i][0];
            bb.y = b.y + dir[i][1];
            if (bb.x >= 0 && bb.x < 3 && bb.y >= 0 && bb.y < 3) {
                bb.s = b.s;
                swap(bb.s[b.x * 3 + b.y], bb.s[bb.x * 3 + bb.y]);
                if (!vis2.count(bb.s)) vis2[bb.s] = dirs[3 - i] + vis2[b.s], q2.push(bb);
            }
        }
    }
    return "unsolvable";
}

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

    node init, ans;
    for (int i = 0;i < 9;i++) {
        char tmp;
        cin >> tmp;
        init.s += tmp;
        if (tmp == 'x') {
            init.x = i / 3;
            init.y = i % 3;
        }
    }
    ans.s = "12345678x";
    ans.x = 2;
    ans.y = 2;
    cout << bfs(init, ans) << '\n';
    return 0;
}