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 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
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.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
算法解析:八数码模板题.
/* BFS+Cantor()模板 解决八数码,十六数码问题. */ //#include<bits/stdc++.h> #include <iostream> #include <algorithm> #include <deque> #include <queue> #include <string> #include <cstring> const int LEN = 362880; using namespace std; struct node{ int state[9]; int dis; char line[1000];//记录状态. }; int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};//上下,右左四个方向 int visited[LEN] = {0};//每个状态对应的记录, 用于标记是否 int start[9];//开始状态 int goal[9]; //目标状态 long int factory[] = {1,1,2,6,24,120,720,5040,40320,362880};//Cantor()用到的常数==> 0!~9! int Cantor(int str[],int n){//用康拓展开来判重 long result = 0;//str排在第几位 for(int i = 0; i < n; i++){ int counted = 0; for(int j = i+1;j < n; j++){ if(str[i]>str[j]){ counted++;//当前未出现的元素排在第几位 } } result+=counted * factory[n-i-1]; } if(!visited[result]){//没有被访问过 visited[result] = 1; return 1; }else{ return 0; } } int bfs(node &desNode){ node head;//用于存放起始点 memcpy(head.state,start,sizeof(head.state)); // memcpy(void *dest, void *src, int count) 由src所指内存区域复制count个字节到dest所指内存区域。 head.dis = 0; queue<node> q; Cantor(head.state,9);//标记起点 已经被访问过. q.push(head);//进队列 while(!q.empty()){ head = q.front(); q.pop(); if(memcmp(head.state,goal,sizeof(goal))==0){ memcpy(desNode.line,head.line,sizeof(head.line)); return head.dis;//到达目标状态,结束 } int z; for(z = 0;z < 9;z++){//找这个状态中元素0的位置 if(head.state[z]==0){ break; } } //转换为二维,左上角是 (0,0) int y = z %3; int x = z / 3; for(int i = 0; i < 4;i++){ int newX = x+dir[i][0];//元素0 转移后的新坐标 int newY = y + dir[i][1]; int nz = newX*3+newY; if(newX>=0&&newX<3&&newY>=0&&newY<3){//判断是否越界 node newnode; memcpy(&newnode,&head,sizeof(struct node));// 用head对newnode进行初始化 swap(newnode.state[z],newnode.state[nz]);// 把0移动到新的位置 newnode.dis++; if(Cantor(newnode.state,9)){//判断是否被访问过 if(i == 0){//上 strcat(newnode.line,"u"); //newnode.line }else if(i==1){//下 strcat(newnode.line,"d"); }else if(i==2){//左 strcat(newnode.line,"l"); }else{//右 strcat(newnode.line,"r"); } q.push(newnode);//将其放入到队列中 } //把0转移到新的位置. } } } return -1;//没找到 } char ch; int main() { node disNode; for(int i = 0;i <9;i++){ cin>>ch; if(ch!='x'){ start[i] = ch - '0'; }else{ start[i] = 0; } //cin>>start[i]; } for(int i = 0;i < 9;i++){ goal[i] = i+1; } goal[8] = 0; int num = bfs(disNode); if(num!=-1){ //printf("%s\n",disNode.line); cout<<disNode.line<<endl; }else{ //printf("unsolvable\n"); cout<<"unsolvable"<<endl; } return 0; }
