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。
这个题很经典，还需要继续做。先把第一次写的代码贴出来吧。
继续优化中
HDU 1043
Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7145 Accepted Submission(s): 1946
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
JGShining
/* HDU 1043 Eight 思路：反向搜索，从目标状态找回状态对应的路径 用康托展开判重 AC G++ 328ms 13924K */ #include<stdio.h> #include<string.h> #include<iostream> #include<queue> #include<string> using namespace std; const int MAXN=1000000;//最多是9!/2 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! bool vis[MAXN];//标记 string path[MAXN];//记录路径 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 num=0; for(int j=i+1;j<9;j++) if(s[j]<s[i])num++; sum+=(num*fac[9-i-1]); } return sum+1; } struct Node { int s[9]; int loc;//“0”的位置 int status;//康拖展开的hash值 string path;//路径 }; int move[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;//123456780对应的康拖展开的hash值 void bfs() { memset(vis,false,sizeof(vis)); 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+move[i][0]; int ty=y+move[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(!vis[next.status]) { vis[next.status]=true; next.path=indexs[i]+next.path; q.push(next); path[next.status]=next.path; } } } } int main() { char ch; Node cur; bfs(); while(cin>>ch) { if(ch=='x') {cur.s[0]=0;cur.loc=0;} else cur.s[0]=ch-'0'; for(int i=1;i<9;i++) { cin>>ch; if(ch=='x') { cur.s[i]=0; cur.loc=i; } else cur.s[i]=ch-'0'; } cur.status=cantor(cur.s); if(vis[cur.status]) { cout<<path[cur.status]<<endl; } else cout<<"unsolvable"<<endl; } return 0; }
POJ 1077
Eight
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 18379 | Accepted: 8178 | 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
/* POJ 1077 Eight 正向广度搜索 把“x"当初0 G++ AC 5200K 719ms */ #include<stdio.h> #include<queue> #include<string.h> #include<iostream> #include<algorithm> 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! bool vis[MAXN];//标记 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 num=0; for(int j=i+1;j<9;j++) if(s[j]<s[i])num++; sum+=(num*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 move[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//u,d,l,r char indexs[5]="udlr";//正向搜索 Node ncur; bool bfs() { memset(vis,false,sizeof(vis)); Node cur,next; queue<Node>q; q.push(ncur); while(!q.empty()) { cur=q.front(); q.pop(); if(cur.status==aim) { path=cur.path; return true; } int x=cur.loc/3; int y=cur.loc%3; for(int i=0;i<4;i++) { int tx=x+move[i][0]; int ty=y+move[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(!vis[next.status]) { vis[next.status]=true; next.path=next.path+indexs[i]; if(next.status==aim) { path=next.path; return true; } q.push(next); } } } return false; } int main() { 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); if(bfs()) { cout<<path<<endl; } else cout<<"unsolvable"<<endl; } return 0; }
这个题目我的做法是把“x"当成0的。
网上很多当成9的话很多不一样了，特意说明下。
康托展开很简单，百度百科上的很容易理解。
谢谢
------------------------------kuangbin