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Eight
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 xwhere 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 8is 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 Source |
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#include<iostream> #include<queue> #include<algorithm> #include<string> #include<cstring> #include<cstdio> //正向广度搜索 //把“x"当初0 using namespace std; const int maxn=1000000; int fac[]={1,1,2,6,24,120,720,5040,40320,362880}; //康拖展开判重 // 0!1!2!3! 4! 5! 6! 7! 8! 9! int vis[maxn]; int Cantor(int s[]){ //康拖展开求该序列的hash值 int sum=0; for(int i=0;i<9;i++){ int cnt=0; for(int j=i+1;j<9;j++) if(s[i]>s[j]) cnt++; sum+=(cnt*fac[9-i-1]); } return sum+1; } struct node{ int s[9]; int loc; //“0”的位置,把“x"当0 int status; //康拖展开的hash值 string path; //路径 }; string path; int aim=46234; //123456780对应的康拖展开的hash值 int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//u,d,l,r char indexs[5]="udlr";//正向搜索 node ncur; int BFS(){ queue<node> q; while(!q.empty()) q.pop(); node cur,next; q.push(ncur); int x,y; while(!q.empty()){ cur=q.front(); q.pop(); if(cur.status==aim){ path=cur.path; return 1; } x=cur.loc/3; y=cur.loc%3; for(int i=0;i<4;i++){ int tx=x+dir[i][0]; int ty=y+dir[i][1]; if(tx<0 || tx>=3 || ty<0 || ty>=3) continue; next=cur; next.loc=tx*3+ty; next.s[cur.loc]=next.s[next.loc]; next.s[next.loc]=0; next.status=Cantor(next.s); if(!vis[next.status]){ vis[next.status]=1; next.path=next.path+indexs[i]; if(next.status==aim){ path=next.path; return 1; } q.push(next); } } } return 0; } int main(){ //freopen("input.txt","r",stdin); char ch; while(cin>>ch){ if(ch=='x'){ ncur.s[0]=0; ncur.loc=0; }else ncur.s[0]=ch-'0'; for(int i=1;i<9;i++){ cin>>ch; if(ch=='x'){ ncur.s[i]=0; ncur.loc=i; }else ncur.s[i]=ch-'0'; } ncur.status=Cantor(ncur.s); memset(vis,0,sizeof(vis)); if(BFS()) cout<<path<<endl; else printf("unsolvable\n"); } return 0; }
HDU 1043 和 POJ 1077 两题类似。。。但是输入不同。
HDU 上是同时多组输入，POJ是单组输入。
两个限时不同。
HDU 上反向搜索，把所有情况打表出来。
POJ上正向搜索。
#include<iostream> #include<queue> #include<string> #include<cstdio> #include<cstring> using namespace std; const int maxn=1000000; int fac[]={1,1,2,6,24,120,720,5040,40320,362880}; int visited[maxn]; string path[maxn]; int Cantor(int s[]){ int sum=0; for(int i=0;i<9;i++){ int cnt=0; for(int j=i+1;j<9;j++) if(s[i]>s[j]) cnt++; sum+=cnt*fac[9-i-1]; } return sum+1; } struct node{ int s[9]; int loc; int status; string path; }ncur; int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//u,d,l,r char indexs[5]="durl";//和上面的要相反，因为是反向搜索 int aim=46234; void BFS() { memset(visited,false,sizeof(visited)); node cur,next; for(int i=0;i<8;i++)cur.s[i]=i+1; cur.s[8]=0; cur.loc=8; cur.status=aim; cur.path=""; queue<node>q; q.push(cur); path[aim]=""; while(!q.empty()) { cur=q.front(); q.pop(); int x=cur.loc/3; int y=cur.loc%3; for(int i=0;i<4;i++) { int tx=x+dir[i][0]; int ty=y+dir[i][1]; if(tx<0||tx>2||ty<0||ty>2)continue; next=cur; next.loc=tx*3+ty; next.s[cur.loc]=next.s[next.loc]; next.s[next.loc]=0; next.status=Cantor(next.s); if(!visited[next.status]) { visited[next.status]=true; next.path=indexs[i]+next.path; q.push(next); path[next.status]=next.path; } } } } int main(){ //freopen("input.txt","r",stdin); char ch; BFS(); while(cin>>ch){ if(ch=='x'){ ncur.s[0]=0; ncur.loc=0; }else ncur.s[0]=ch-'0'; for(int i=1;i<9;i++){ cin>>ch; if(ch=='x'){ ncur.s[i]=0; ncur.loc=i; }else ncur.s[i]=ch-'0'; } ncur.status=Cantor(ncur.s); if(visited[ncur.status]) cout<<path[ncur.status]<<endl; else cout<<"unsolvable"<<endl; } return 0; }