Eight hdu 1043 poj 1077

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 Input

2 3 4 1 5 x 7 6 8

题意 ： 就是8数码问题，主要时搜索的路径寻找问题，把'x'转换为0，然后把当前这些数都存储为1个状态

分析： 这道题有两道相同的，分别是HDU和POJ,HDU上的数据加强了比POJ更麻烦。。

ＰＯＪ：正向搜索就好（简单）

 #include <iostream>
 #include <cstring>  

 using namespace std;  

 const int maxn = ;
 typedef int State[];
 State st[maxn];
 int goal[] = {, , , , , , , , };
 int dx[] = {-, ,  , };
 int dy[] = { , , -, };
 int head[maxn], nxt[maxn], fa[maxn];
 char dir[maxn];  

 int Hash(State s)       //哈希函数
 {
     int ret = , i;
     for(i = ; i < ; i++) ret = ret *  + s[i];
     return ret % maxn;
 }  

 bool try_to_insert(int rear)        //插入哈希表
 {
     int h = Hash(st[rear]);
     for(int e = head[h]; e != -; e = nxt[e])
     {
         if(memcmp(st[e], st[rear], sizeof(st[e])) == ) return ;
     }
     nxt[rear] = head[h];
     head[h] = rear;
     return ;
 }  

 int bfs()       //遍历
 {
     int frt = , rear = , i, z;
     while(frt < rear)
     {
         State& s = st[frt];
         if(memcmp(s, goal, sizeof(s)) == ) return frt;
         for(z = ; s[z] != ; z++);
         int x = z / ;
         int y = z % ;
         for(i = ; i < ; i++)
         {
             int newx = x + dx[i];
             int newy = y + dy[i];
             int newz =  * newx + newy;
             if(newx >=  && newx <  && newy >=  && newy < )
             {
                 State& news = st[rear];
                 memcpy(news, s, sizeof(s));
                 news[z] = s[newz];
                 news[newz] = ;
                 if(try_to_insert(rear))  　　　//注意这里的路径输出的方式
                 {
                     fa[rear] = frt;
                     switch(i)
                     {
                         case : dir[rear] = 'u'; break;
                         case : dir[rear] = 'd'; break;
                         case : dir[rear] = 'l'; break;
                         case : dir[rear] = 'r'; break;
                         default: break;
                     }
                     rear++;
                 }
             }
         }
         frt++;
     }
     return ;
 }  

 void print(int i)       //输出
 {
     if(fa[i] == -) return;
     print(fa[i]);
     cout<<dir[i];
 }  

 int main()
 {
     char c[];
     int i, ret;
     while(cin>>c[]>>c[]>>c[]>>c[]>>c[]>>c[]>>c[]>>c[]>>c[])
     {
         for(i = ; i < ; i++) st[][i] = c[i] == 'x' ?  : (int)(c[i]-'');
         memset(head, -, sizeof(head));
         fa[] = -;
         ret = bfs();
         if(ret)
         {
             print(ret);
         }
         else cout<<"unsolvable";
         cout<<endl;
     }
     return ;
 }

HDU : 这道题时多组输入，所以不能向上面一样在线写，而是要从最终状态开始倒着把所有状态搜索一遍，之后只需要输入初始状态打表判断输出路径即可；

　　学习到的知识有两个：bfs()路径查找类　＋　康拓展开，路径的输出：

 /*************************************************************************
     > File Name: search.cpp
     > Author : PrayG
     > Mail: 996930051@qq,com
     > Created Time: 2016年07月20日 星期三 10时56分09秒
  ************************************************************************/

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<queue>
 #include<algorithm>
 #include<map>
 #include<stack>
 #include<set>
 #include<cmath>
 using namespace std;
 const int maxn = ;
 int fac[] = {,,,,,,,,,};
 int dx[] = {,,-,},dy[] = {,,,-};//drul
 char ind[] = "uldr";//与上面相反
 string path[maxn];//记录路径
 bool vis[maxn];
 int aim = ;//123456780 的康拓展开

 struct node
 {
     int s[];  //记录状态
     int sit0;　　//0 的位置
     int val;　　　//康拓展开的值
     string path;　　// 路径
 };

 int cant(int s[])　　//康拓展开
 {
     int code = ;
     for(int i = ; i < ; i++)
     {
         int cnt = ;
         for(int  j= i+ ; j < ; j++)
         {
             if(s[i] > s[j])
             {
                 cnt++;
             }
         }
         code += fac[-i] * cnt;
     }
     return code;
 }

 void bfs()
 {
     memset(vis,false,sizeof(vis));
     queue<node> que;
     node cnt1,cnt2;
     for(int i =  ; i < ;i++)
       cnt1.s[i] = i+;
     cnt1.s[] = ;
     cnt1.sit0 = ;
     //printf("aim = %d\n",aim);
     cnt1.val = aim;
     cnt1.path = "";
     path[aim] = "";
     que.push(cnt1);
     while(!que.empty())
     {
         cnt1 = que.front();
         que.pop();
         int x = cnt1.sit0 / ;
         int y = cnt1.sit0 % ;
         for(int i = ; i < ; i++)
         {
             int nx = x + dx[i];
             int ny = y + dy[i];
             int nz = nx *  + ny;
             if(nx <  || nx > || ny < || ny >)
                continue;
             cnt2 = cnt1;
             cnt2.s[cnt1.sit0] = cnt2.s[nz];
             cnt2.s[nz] = ;
             cnt2.sit0 = nz;
             cnt2.val = cant(cnt2.s);
             if(!vis[cnt2.val])
             {
                 vis[cnt2.val] = true;
                 cnt2.path = ind[i] + cnt1.path;
                 que.push(cnt2);
                 path[cnt2.val] = cnt2.path;
             }
         }

     }
 }

 int main()
 {
     bfs();
     char t;
     while(cin >> t)
     {
         node st;
         if(t == 'x'){
           st.s[] = ;
             st.sit0 = ;
         }
         else
         st.s[] = t - '';
         for(int i = ; i< ; i++)
         {
             cin >> t;
             if(t == 'x')
             {
                 st.s[i] = ;
                 st.sit0 = i;
             }
             else
              st.s[i] = t -'';
         }
         st.val = cant(st.s);
         if(vis[st.val])
         {
             cout << path[st.val] << endl;
         }
         else
          cout << "unsolvable" << endl;
     }
     return ;
 }