首页 > 代码库 > HDU - 1043 - Eight / POJ - 1077 - Eight
HDU - 1043 - Eight / POJ - 1077 - Eight
先上题目：
Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11243 Accepted Submission(s): 3022
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
题意：经典的八数码，给出当前状态，求出恢复到从小到大排列的顺序的任意一种路径，如果不存在就输出unsolvable。这一题用搜索解决。在看到其他人的题解看到这一题有很多种解法，这里我尝试的是用A*，第一次用A*→_→。
先说说解法，这一题需要先知道的知识：①对于八数码有解的情况，用当前的八个数字组成的序列，如果逆序对的对数是偶数的话就有解，否则无解，这是判断八数码有没有解的条件。②这里如果选择使用哈希记录状态的话，需要想个不错的方法。这里使用的方法是用康托展开。对于这东西可以自行百度一下。
这里使用A*，需要注意的是，我们选择扩展的状态是选择f(s)(估值函数)最小的状态。其中在计算g(s)(已使用代价)用的是每扩展一次就加一，h(s)(预估使用代价)是使用每一个位置到达正确位置的哈密顿距离之和来作为预估值。在将状态放入优先队列(OPEN表)的时候根据不同顺序比较g(s),h(s)效果可能不一样。
A*的本质就是每一次从还没有扩展的状态集合里面选出f(s)最小的那个状态来扩展，感觉看起来就是优先队列版的BFS。
当然，A*不一定就能提高效率，这就得看估值函数的选取了。
这一题码了两次，第一次是基本上跟着别人的代码写的，第二次自己写，但还是不得不稍微看一些自己写的代码，主要是对康托展开不是很熟悉。
上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 #include <algorithm> 6 #define MAX 400002 7 using namespace std; 8 typedef struct Node{ 9 int maze[3][3]; 10 int y,x; 11 int hash; 12 int h,g; 13 bool isok(){ 14 return (0<=y && y<3 && 0<=x && x<3); 15 } 16 17 bool operator < (const Node a) const{ 18 if(h>a.h) return 1; 19 if(h==a.h && g>a.g) return 1; 20 return 0; 21 } 22 23 }Node; 24 Node s; 25 const int cantor[9] = {1,1,2,6,24,120,720,5040,40320}; 26 const int target = 322560; 27 const int cy[]={0,0,1,-1}; 28 const int cx[]={1,-1,0,0}; 29 int vis[MAX]; 30 int pre[MAX]; 31 priority_queue<Node> p; 32 bool check(Node c){ 33 int a[9]; 34 int k=0,sum=0; 35 for(int i=0;i<3;i++) for(int j=0;j<3;j++) a[k++]=c.maze[i][j]; 36 for(int i=0;i<k;i++) for(int j=i+1;j<k;j++) if(a[i] && a[j] && a[i]>a[j]) sum++; 37 return !(sum&1); 38 } 39 40 int getHash(Node c){ 41 int a[9]; 42 int k=0,h=0,sum=0; 43 for(int i=0;i<3;i++) for(int j=0;j<3;j++) a[k++]=c.maze[i][j]; 44 for(int i=0;i<k;i++){ 45 sum=0; 46 for(int j=0;j<i;j++){ 47 if(a[i]<a[j]) sum++; 48 } 49 h+=sum*cantor[i]; 50 } 51 return h; 52 } 53 54 int getH(Node c){ 55 int res=0; 56 for(int i=0;i<3;i++) for(int j=0;j<3;j++){ 57 res+=abs(i-(c.maze[i][j]-1)/3)+abs(j-(c.maze[i][j]-1)%3); 58 } 59 return res; 60 } 61 62 void astar(){ 63 while(!p.empty()) p.pop(); 64 memset(vis,-1,sizeof(vis)); 65 memset(pre,-1,sizeof(pre)); 66 Node u,v; 67 s.h = getH(s); 68 p.push(s); 69 vis[s.hash]=-2; 70 while(!p.empty()){ 71 u = p.top(); 72 p.pop(); 73 for(int i=0;i<4;i++){ 74 v=u;v.y+=cy[i];v.x+=cx[i]; 75 if(!v.isok()) continue; 76 swap(v.maze[v.y][v.x],v.maze[u.y][u.x]); 77 if(!check(v)) continue; 78 v.hash = getHash(v); 79 if(vis[v.hash]!=-1) continue; 80 vis[v.hash]=i; pre[v.hash]=u.hash; 81 v.g++; 82 v.h=getH(v); 83 p.push(v); 84 if(v.hash == target) return ; 85 86 } 87 } 88 89 } 90 91 bool scan(){ 92 char ch[100]; 93 if(!gets(ch)) return 0; 94 int k=0,l=strlen(ch); 95 for(int i=0;i<3;i++) for(int j=0;j<3;j++){ 96 while(k<l && ch[k]==‘ ‘) k++; 97 if(ch[k]==‘x‘){ 98 s.maze[i][j]=0; 99 s.y=i; s.x=j;100 }101 else s.maze[i][j]=ch[k]-‘0‘;102 k++;103 }104 return 1;105 }106 107 void printPath(){108 string ss;109 ss.clear();110 int next = target;111 while(pre[next]!=-1){112 if(vis[next]==0) ss+=‘r‘;113 else if(vis[next]==1) ss+=‘l‘;114 else if(vis[next]==2) ss+=‘d‘;115 else ss+=‘u‘;116 next = pre[next];117 }118 for(int i=(int)ss.length()-1;i>=0;i--) putchar(ss[i]);119 puts("");120 }121 122 int main()123 {124 //freopen("data.txt","r",stdin);125 while(scan()){126 if(!check(s)){127 puts("unsolvable");128 }else{129 s.hash = getHash(s);130 if(s.hash == target){131 puts("");132 }else{133 astar();134 printPath();135 }136 }137 }138 return 0;139 }
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <queue> 6 #define MAX 400002 7 using namespace std; 8 9 const int cantor[] = {1,1,2,6,24,120,720,5040,40320}; 10 const int target = 322560; 11 int vis[MAX],pre[MAX]; 12 int cy[]={0,1,0,-1}; 13 int cx[]={1,0,-1,0}; 14 typedef struct Node{ 15 int maze[3][3]; 16 int h,g,y,x; 17 int hash; 18 19 bool isok(){ 20 return (0<=y && y<3 && 0<=x && x<3); 21 } 22 23 bool islegal(){ 24 int a[9],k=0,sum=0; 25 for(int i=0;i<3;i++) for(int j=0;j<3;j++) a[k++]=maze[i][j]; 26 for(int i=0;i<k;i++) for(int j=i+1;j<k;j++) if(a[i] && a[j] && a[i]>a[j]) sum++; 27 return !(sum&1); 28 } 29 30 bool operator < (const Node o)const{ 31 return h==o.h ? g>o.g : h>o.h; 32 } 33 }Node; 34 Node s; 35 priority_queue<Node> q; 36 37 int getHash(Node e){ 38 int a[9],k=0,res=0; 39 for(int i=0;i<3;i++) for(int j=0;j<3;j++) a[k++]=e.maze[i][j]; 40 for(int i=0;i<k;i++){ 41 int t=0; 42 for(int j=0;j<i;j++) if(a[i]<a[j]) t++; 43 res+=t*cantor[i]; 44 } 45 return res; 46 } 47 48 int getH(Node e){ 49 int res=0; 50 for(int i=0;i<3;i++) for(int j=0;j<3;j++){ 51 res+=abs(i-(e.maze[i][j]-1)/3)+abs(j-(e.maze[i][j]-1)%3); 52 } 53 return res; 54 } 55 56 void astar(){ 57 Node u,v; 58 while(!q.empty()) q.pop(); 59 memset(vis,-1,sizeof(vis)); 60 memset(pre,-1,sizeof(pre)); 61 s.g=0; 62 s.h=getH(s); 63 q.push(s); 64 vis[s.hash]=-2; 65 while(!q.empty()){ 66 u=q.top(); 67 q.pop(); 68 //cout<<u.hash<<endl; 69 for(int i=0;i<4;i++){ 70 v=u;v.y+=cy[i];v.x+=cx[i]; 71 if(v.isok()){ 72 swap(v.maze[v.y][v.x],v.maze[u.y][u.x]); 73 if(v.islegal()){ 74 v.hash = getHash(v); 75 if(vis[v.hash]==-1){ 76 vis[v.hash]=i; 77 pre[v.hash]=u.hash; 78 v.g++; v.h=getH(v); 79 q.push(v); 80 if(v.hash == target) return ; 81 } 82 } 83 84 } 85 } 86 } 87 } 88 89 void printPath(){ 90 string ss=""; 91 int next=target; 92 while(pre[next]!=-1){ 93 switch(vis[next]){ 94 case 0:ss+=‘r‘;break; 95 case 1:ss+=‘d‘;break; 96 case 2:ss+=‘l‘;break; 97 case 3:ss+=‘u‘;break; 98 } 99 next=pre[next];100 }101 for(int i=(int)ss.length()-1;i>=0;i--) putchar(ss[i]);102 putchar(‘\n‘);103 }104 105 bool get(){106 char ss[100];107 if(gets(ss)){108 int k=0,l=strlen(ss);109 for(int i=0;i<3;i++) for(int j=0;j<3;j++){110 while(k<l && ss[k]==‘ ‘) k++;111 if(ss[k]==‘x‘){112 s.maze[i][j]=0;113 s.y=i; s.x=j;114 }else s.maze[i][j]=ss[k]-‘0‘;115 k++;116 }117 return 1;118 }119 return 0;120 121 }122 123 int main()124 {125 //freopen("data.txt","r",stdin);126 while(get()){127 if(s.islegal()){128 s.hash = getHash(s);129 if(s.hash == target) puts("");130 else{131 astar();132 printPath();133 }134 }else puts("unsolvable");135 }136 return 0;137 }
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