Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10101 Accepted Submission(s): 2684
Special Judge
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:
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:
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.
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
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
#include <iostream> #include <string> #include <string.h> #include <map> #include <stdio.h> #include <algorithm> #include <queue> #include <vector> #include <math.h> #include <set> #define Max(a,b) ((a)>(b)?(a):(b)) #pragma comment(linker, "/STACK:16777216") using namespace std ; typedef long long LL ; const int fac[9] = {1,1,2,6,24,120,720,5040,40320} ; const int d[4][2] = {{1,0},{-1,0},{0,1},{0,-1}} ; //四个方向 const char direction[4] = {'d','u','r','l'} ; //输出记录 const int end_pos[10][2] = {{0,0},{0,0},{0,1},{0,2}, {1,0},{1,1},{1,2}, {2,0},{2,1},{2,2}} ; //最后状态 123 // 456 // 78X class Node{ public : char mat[3][3] ; int x ; int y ; int h ; //h(x)=递归层数 int g ; //g(x)=曼哈顿距离和 int f ; //估价函数 f(x) = g(x)+ h(x) int id ; //hash值,用于判重,利用排列序列的康拓展开 int G() ;//求g（x） int cango() ; //剪枝，为什么逆序数必须是偶数，网上很多人问这个问题。 //我的解答： 与X（空白）交换的Y ，编号（index）相差为偶数，即在[ index[X] + 1 .index[Y]-1 ] 区间的数为偶数 ； // 交换前 ，设 [ index[X] + 1 .index[Y]-1 ](zh) ，〉Y有a个，<Y 有b个 ，则a+b为偶数 ； // 交换后 ，逆序数相较之前 +b -a ,即| 逆序数交换前 -逆序数交换后 | 为偶数 。 // 也就是说无论何种交换，逆序数的变化差值为偶数 ，而最终的状态12345678X 。 (X可看作为9) ，逆序数为0。 // 即只能从逆序数为偶数的状态转移 。 int Hash() ; // 求id void out() ; Node(){}; Node(vector<char>) ; friend bool operator < (const Node &A ,const Node &B){ if(A.f != B.f) return A.f > B.f ; else return A.g > B.g ; } //建堆，使用C++ STL 优先队列 ，f(x)=g(x)+h（x） , f(x)小的优先级别大，也就是说与最后的状态差异小的优先考虑 }; Node::Node(vector<char>s){ int i , j ; for(int k = 0 ;k < s.size() ;k++){ i = k/3 ; j = k%3 ; if(s[k] == 'x'){ this->x = i ; this->y = j ; } mat[i][j] = s[k] ; } } int Node::G(){ int sum = 0 ; for(int i = 0 ;i <= 2 ;i++){ for(int j = 0 ;j <= 2 ;j++){ if(mat[i][j]=='x') continue ; int n = mat[i][j] - '0' ; sum = sum + abs(i-end_pos[n][0]) + abs(j - end_pos[n][1]) ; } } return sum ; } int Node::cango(){ char num[10] ; int n = 0 ,sum = 0; for(int i = 0 ;i <=2 ;i++){ for(int j = 0 ;j <= 2 ;j++){ if(mat[i][j] != 'x') num[++n] = mat[i][j] ; } } for(int i = 1 ;i <= n ;i++){ for(int j =1 ;j < i ;j++){ if(num[j] > num[i]) sum++ ; } } return (sum&1) == 0 ; } int Node::Hash(){ char num[10] ; int n = 0 ,sum ,index = 0 ; for(int i = 0 ;i <=2 ;i++){ for(int j = 0 ;j <= 2 ;j++) num[n++] = mat[i][j] ; } for(int i = 0 ;i < n ;i++){ sum = 0 ; for(int j =0 ;j < i ;j++){ if(num[j] > num[i]) sum++ ; } index += fac[i]*sum ; } return index ; } void Node::out(){ for(int i = 0 ;i <= 2 ;i++){ for(int j = 0 ;j <= 2 ;j++) putchar(mat[i][j]) ; puts("") ; } } int father[363000] ; class App{ private : bool visited[363000] ; Node start ; char opet[363000] ; public : App(){} App(class Node) ; int cango(int ,int) ; int A_star() ; void out() ; }; App::App(class Node s){ start = s ; start.h = 0 ; start.g = start.G() ; start.f = start.g + start.h ; start.id = start.Hash() ; memset(visited,0,sizeof(visited)) ; visited[start.id] = 1 ; } int App::cango(int x ,int y){ return 0<=x&&x<=2&&0<=y&&y<=2 ; } int App::A_star(){ if(start.id == 0) return 1 ; if(!start.cango()) return 0; priority_queue<Node>que ; que.push(start) ; while(!que.empty()){ Node now =que.top() ; que.pop() ; if(now.id == 0) return 1 ; for(int i = 0 ;i < 4 ;i++){ int x = now.x + d[i][0] ; int y = now.y + d[i][1] ; if(!cango(x,y)) continue ; Node next = now ; next.x = x ; next.y = y ; swap(next.mat[now.x][now.y],next.mat[next.x][next.y]) ; next.id = next.Hash() ; if(visited[next.id]) continue ; visited[next.id] = 1 ; if(!next.cango()) continue ; next.g = next.G() ; next.h = now.h +1 ; next.f = next.g + next.h ; father[next.id] = now.id ; opet[next.id] = direction[i] ; que.push(next) ; } } return 0 ; } void App::out(){ if(A_star()==0) puts("unsolvable") ; else{ vector<char>ans ; ans.clear() ; int i = 0 ; while(start.id != i){ ans.push_back(opet[i]) ; i = father[i] ; } for(int i = ans.size()-1 ;i >= 0 ;i--) putchar(ans[i]) ; puts("") ; } } int main(){ char str[20] ; vector<char> s ; while(gets(str)){ s.clear() ; for(int i = 0 ;i < strlen(str) ;i++){ if(str[i] != ' ') s.push_back(str[i]) ; } Node start ; start = Node(s) ; App app = App(start) ; app.out() ; } return 0 ; }
相关知识点：
A*算法
公式表示为： f(n)=g(n)+h(n),
其中 f(n) 是从初始点经由节点n到目标点的估价函数，
g(n) 是在状态空间中从初始节点到n节点的实际代价，
h(n) 是从n到目标节点最佳路径的估计代价。
保证找到最短路径（最优解的）条件，关键在于估价函数h(n)的选取：
估价值h(n)<= n到目标节点的距离实际值，这种情况下，搜索的点数多，搜索范围大，效率低。但能得到最优解。
如果 估价值>实际值,搜索的点数少，搜索范围小，效率高，但不能保证得到最优解。
康托展开
{1,2,3,4,...,n}表示1,2,3,...,n的排列如 {1,2,3} 按从小到大排列一共6个 123 132 213 231 312 321
代表的数字 1 2 3 4 5 6 也就是把10进制数与一个排列对应起来。他们间的对应关系可由康托展开来找到。 如我想知道321是{1,2,3}中第几个大的数可以这样考虑 第一位是3，当第一位的数小于3时，那排列数小于321 如 123 213 小于3的数有1，2 所以有2*2!个 再看小于第二位2的 小于2的数只有一个就是1 所以有1*1!=1 所以小于321的{1,2,3}排列数有2*2!+1*1!=5个所以321是第6个大的数。 2*2!+1*1!是康托展开 再举个例子 1324是{1,2,3,4}排列数中第几个大的数 第一位是1小于1的数没有，是0个 0*3! 第二位是3小于3的数有1，2但1已经在第一位了所以只有一个数2 1*2! 第三位是2小于2的数是1，但1在第一位所以有0个数 0*1! 所以比1324小的排列有0*3!+1*2!+0*1!=2个 1324是第三个大数。