链接:https://ac.nowcoder.com/acm/problem/51032
来源:牛客网

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.
注意:题目输出不是唯一的,即这ulrd四个字母没有优先级!如果你的代码写错了,并不是优先级的问题,而是其他地方写错了!
还有就是记得无解时输出"unsolvable"(不含引号)

思路

看到这个数据范围是3*3(相当于n==9),再看到题目每次移动相当于从上下左右四个方向走一遍-->BFS求解
(如果不懂BFS的大致模板,请自学)
来看看判断条件
1.BFS结束执行的条件,即变成“123456789”的样子(本来是“12345678x”的,但是这里我们发现,12345678是数字,x是字符,所以可以把x读入时变成9,相当于把字符串比较转化成了数字比较,相对方便一些
2.每次BFS搜索的位置:上下左右,上为u,下为d,左为l,右为r,依次考虑即可
3.每次BFS后是否要回溯原来的点:是,因为部分格子存在重复经过的情况
4.是否需要剪枝:是,因为可能会出现循环移动的情况(比如说右下左上)
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明白了这些判断条件,我们看看,如何将这些条件转化成代码(可以结合下面的代码食用)
先看第个条件,很简单,先开一个方向数组(代码中的dir数组),然后for循环每个方向就行,记得判断位置坐标是否合法
再看第个条件,由于在队列中每一次都会存储当前的mp的状态,所以每次判断是否结束时,直接把这个状态拿出来,与123456789直接比较是否相等即可,只有当某一次相等了才说明有解。
如果队列中的状态已经全部遍历完了还没有解,那么必定无解。
再看第个条件,由于一开始输入的数字存在了mp数组里,相当于一张图,所以需要将图中的指定坐标交换(即相当于执行移动操作)之后,先转化成数字串(不是字符串!)的形式,方便存入队列中;之后拿出来使用时,再转化成图的形式即可。
下面的代码中的zh()函数即为图到数字串的转化,zhn()函数即为数字串到图的转化
最后看剪枝,这里其实用map或者set都行(因为set中的元素不会重复嘛),下面代码里用的是map,大致思路为:每次看看map中已经存储的数字,是否为swap()之后的新数字,不是的话,就可以将这个数字存入队列中,是的话就不存,这样剪枝,有效的避免了死循环的发生。
好了,差不多分析完了,可以开始写代码了

AC代码如下

(与其他题解不同,没有哈希,unordered_map是<int,int>型的)

#include<bits/stdc++.h>
using namespace std;
struct node {
    int num,x,y;
    string move;
};
queue<node>q;
unordered_map<int,int>d;
int mp[4][4];
int dir[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
string ans;
int zh() {
    int sum=0;
    for(int i=1; i<=3; i++) {
        for(int j=1; j<=3; j++) {
            sum=sum*10+mp[i][j];
        }
    }
    return sum;
}
void zhn(int n){
    int sum=n;
    for(int i=3;i>=1;i--){
        for(int j=3;j>=1;j--){
            mp[i][j]=(sum%10);
            sum/=10;
        }
    }
}
string BFS() {
    while(!q.empty()) {
        node tmp=q.front();
        q.pop();
        if(tmp.num==123456789) {
            return tmp.move;
        }
        zhn(tmp.num);
        for(int i=0; i<4; i++) {
            int xx=tmp.x+dir[i][0];
            int yy=tmp.y+dir[i][1];
            if(xx<1 || xx>3 || yy<1 || yy>3) continue;
            swap(mp[tmp.x][tmp.y],mp[xx][yy]);
            int nowtmp=zh();
            if(!d.count(nowtmp)) {
                d[nowtmp]++;
                if(i==0) q.push(node {nowtmp,xx,yy,tmp.move+"d"});
                if(i==1) q.push(node {nowtmp,xx,yy,tmp.move+"r"});
                if(i==2) q.push(node {nowtmp,xx,yy,tmp.move+"u"});
                if(i==3) q.push(node {nowtmp,xx,yy,tmp.move+"l"});
            }
            swap(mp[tmp.x][tmp.y],mp[xx][yy]);
        }
    }
    cout<<"unsolvable"<<endl;
    return " ";
}
int main() {
    for(int i=1; i<=3; i++) {
        for(int j=1; j<=3; j++) {
            char tmp;
            cin>>tmp;
            if(tmp=='x') mp[i][j]=9;
            else mp[i][j]=(tmp-48);
        }
    }
    for(int i=1; i<=3; i++) {
        for(int j=1; j<=3; j++) {
            if(mp[i][j]==9) {
                int sumtt=zh();
                q.push(node {sumtt,i,j,""});
                ans=BFS();
            }
        }
    }
    if(ans!=" ") cout<<ans<<endl;
    return 0;
}