首页 > 代码库 > POJ 1077 Eight（康托展开+BFS）
POJ 1077 Eight（康托展开+BFS）
Eight
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 30176 | Accepted: 13119 | Special Judge | ||
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 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
is described by this list:
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
Source
South Central USA 1998
题目链接：POJ 1077
主要就是用康托展开来映射判重的问题，info::val就是康托展开hash值，info::step就是积累的状态。另外感觉这题剧毒，自己本来用int vis[]和char his[]想最后回溯记录答案从而代替速度比较慢的string，结果居然超时……TLE一晚上，要不是看了大牛的博客估计要一直T在这个坑点上。还有不知道为什么string的加号重载在C++编译器里会CE，换G++才过。
什么是康托展开？——康托展开介绍文章
代码：
#include<iostream>#include<algorithm>#include<cstdlib>#include<sstream>#include<cstring>#include<bitset>#include<cstdio>#include<string>#include<deque>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>using namespace std;#define INF 0x3f3f3f3f#define CLR(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair<int,int> pii;typedef long long LL;const double PI=acos(-1.0);const int N=362880+20;int fact[10]={1,1,2,6,24,120,720,5040,40320,362880};int direct[4][2]={{-1,0},{1,0},{0,-1},{0,1}};char MOVE[5]="udlr";struct info{ int s[9]; int indx; string step; int val;};info S,E;int T;int vis[N];string ans;int calcantor(int s[]){ int r=0; for (int i=0; i<9; ++i) { int k=0; for (int j=i+1; j<9; ++j) { if(s[j]<s[i]) ++k; } r=r+k*fact[8-i]; } return r;}bool check(const int &x,const int &y){ return (x>=0&&x<3&&y>=0&&y<3);}bool bfs(){ CLR(vis,0); queue<info>Q; Q.push(S); vis[S.val]=1; info now,v; while (!Q.empty()) { now=Q.front(); Q.pop(); if(now.val==T) { ans=now.step; return true; } for (int i=0; i<4; ++i) { int x=now.indx/3; int y=now.indx%3; x+=direct[i][0]; y+=direct[i][1]; if(check(x,y)) { v=now; v.indx=x*3+y; v.s[now.indx]=v.s[v.indx]; v.s[v.indx]=0; v.val=calcantor(v.s); if(!vis[v.val]) { vis[v.val]=1; v.step=now.step+MOVE[i]; if(v.val==T) { ans=v.step; return true; } Q.push(v); } } } } return false;}int main(void){ char temp; int i; T=46233; while (cin>>temp) { if(temp==‘x‘) { S.s[0]=0; S.indx=0; } else S.s[0]=temp-‘0‘; for (i=1; i<9; ++i) { cin>>temp; if(temp==‘x‘) { S.s[i]=0; S.indx=i; } else S.s[i]=temp-‘0‘; } S.val=calcantor(S.s); puts(!bfs()?"unsolvable":ans.c_str()); } return 0;}POJ 1077 Eight（康托展开+BFS）
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。