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