Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10351 Accepted Submission(s): 2755
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
Source
Recommend
题意:
就是还原八数码。输出操作。
思路:
利用康拓展开判重。通过康拓展开知一共有362879种状态。用逆序数判断可行解见八数码可行解。然后宽搜。
详细见代码:
#include<algorithm> #include<iostream> #include<string.h> #include<sstream> #include<stdio.h> #include<math.h> #include<vector> #include<string> #include<queue> #include<set> #include<map> //#pragma comment(linker,"/STACK:1024000000,1024000000") using namespace std; const int INF=0x3f3f3f3f; const double eps=1e-8; const double PI=acos(-1.0); const int maxn=370010;//最大为362879 const int num=9; const int over=46233; typedef __int64 ll; int c[10],head,tail; int vis[maxn]; int fac[15]={1,1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600};//存阶乘 int dx[4]={-1,1,-3,3};//四个方向 map<int,char> mpp; struct yb { int mp[10]; int bp,pre,code,dre;//空格位置。上一状态的下标(记录路径),康拓展开,方向。 } q[maxn],tp,tt; /***********树状数组求逆序数(规模比较小快不了多少)***********/ int lowbit(int x) { return x&-x; } void update(int x) { while(x<=num) { c[x]++; x+=lowbit(x); } } int getsum(int x) { int sum=0; while(x>0) { sum+=c[x]; x-=lowbit(x); } return sum; } /***************康拓展开***************/ ll contor(int arr[]) { int i,j,ct; ll sum=0; for(i=0;i<num;i++) { ct=0; for(j=i+1;j<num;j++) if(arr[j]<arr[i]) ct++; sum+=ct*fac[num-i-1]; } return sum; } bool ok(int px,int x,int dre)//判断是否越界 { int lim=((px+3)/3)*3-1; if(x<0||x>=num) return false; if(dre>1) return true; if(x>lim||x<lim-2) return false; return true; } void bfs() { int i,x; memset(vis,0,sizeof vis); vis[tp.code=contor(tp.mp)]=1; tp.pre=-1; head=tail=0; q[tail++]=tp; while(head<tail) { tp=q[head]; if(tp.code==over) return; for(i=0;i<4;i++) { x=tp.bp+dx[i]; if(ok(tp.bp,x,i)) { tt=tp; tt.mp[tt.bp]=tt.mp[x]; tt.mp[x]=0; tt.code=contor(tt.mp); if(!vis[tt.code]) { vis[tt.code]=1; tt.bp=x; tt.pre=head; tt.dre=i; q[tail++]=tt; } } } head++; } } void print(int x)//输出路径 { if(q[x].pre==-1) { //show(q[x].mp); return; } print(q[x].pre); printf("%c",mpp[q[x].dre]); //show(q[x].mp); } int main() { int i,ni,ct; char com[10]; mpp.clear(); mpp[0]=‘l‘;//建立方向和下标的映射 mpp[1]=‘r‘; mpp[2]=‘u‘; mpp[3]=‘d‘; while(~scanf("%s",com)) { ct=0; if(com[0]!=‘x‘) tp.mp[0]=com[0]-‘0‘; else tp.mp[0]=0,tp.bp=0; for(i=1;i<9;i++) { scanf("%s",com); if(com[0]!=‘x‘) tp.mp[i]=com[0]-‘0‘; else tp.mp[i]=0,tp.bp=i; } for(i=0;i<9;i++) if(tp.mp[i]) tt.mp[ct++]=tp.mp[i]; ni=0; memset(c,0,sizeof c); for(i=0;i<8;i++)//求逆序数 { update(tt.mp[i]); ni+=i+1-getsum(tt.mp[i]); } if(ni&1) { printf("unsolvable\n"); continue; } bfs(); print(head); printf("\n"); } return 0; } /******************debug********************/ void show(int arr[]) { int i,j; for(i=0;i<3;i++) { for(j=0;j<3;j++) { printf("%d ",arr[i*3+j]); } printf("\n"); } printf("--------------------\n"); getchar(); }