Eight HDU - 1043
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Eight HDU - 1043
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 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
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
题目大意：给你一个3*3的矩阵，有数字1~8和一个x，问最少进行多少次移动能变成如这样
然后把移动的过程数出出来。
思路：因为有很多个样例，如果每次从所给样例搜，会浪费很多时间，为了把时间减少，我们可以倒着搜，从最后的情况看能搜到什么情况，并把它标记起来，最后在输出路径，大大减少了时间。然后还得用到map容器，①这是因为用book数组首先不能开到九位数那么大，他可以代替book起标记作用，②其次，他可以通过键值来找到不同类型的已经指定类型的映照数据的值，很方便。然后就是基本的广搜，通过已知倒推，并记录路径。还有就是一开始的处理，把x当成9或0，我帮他当成了9来处理。
map映照容器：他是有一个键值和一个映照数据组成的，键值和映照数据之间具有一一映照的关系。键值不允许重复，比较函数只对元素的键值进行比较，元素各项数据可通过键值检索出来。创建map对象，键值和映照数据的类型由自己定义。在没有制定比较函数时，元素的插入位置是按键值有大到小插入到黑白书中。
代码如下：
#include<stdio.h>//没加<string>
#include<queue>
#include<map>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int dir[4][2]={0,1,1,0,0,-1,-1,0};
map<ll,string>m;
map<ll,int>n;
struct node
{
string c;
int s[3][3];
int x,y;
};
void bfs()
{
queue<node>Q;
node p,q;
int k=1;
for(int i=0;i<3;i++)
for(int j=0;j<3;j++)
p.s[i][j]=k++;
p.x=2;
p.y=2;
p.c="";//加上
n[123456789]=1;
Q.push(p);
while(!Q.empty())
{
p=Q.front();
Q.pop();
for(int i=0;i<4;i++)
{
int tx=p.x+dir[i][0];
int ty=p.y+dir[i][1];
if(tx<0||tx>2||ty<0||ty>2)
continue;
q=p;//改了一下
swap(q.s[tx][ty],q.s[p.x][p.y]);
ll sum=0;
for(int j=0;j<3;j++)
for(int k=0;k<3;k++)
sum=sum*10+q.s[j][k];
if(n[sum])
continue;
n[sum]=1;
if(i==0)//注意这里要倒着写
q.c=p.c+'l';
if(i==1)
q.c=p.c+'u';
if(i==2)
q.c=p.c+'r';
if(i==3)
q.c=p.c+'d';
q.x=tx;
q.y=ty;
Q.push(q);
m[sum]=q.c;
}
}
}
int main()
{
bfs();
char a[50];
while(gets(a))
{
ll sum=0;
for(int i=0;i<strlen(a);i++)
{
if(a[i]==' ')
continue;
if(a[i]=='x')
a[i]='9';
sum=sum*10+a[i]-48;
}
if(!n[sum])
printf("unsolvable\n");
else
{
string ss=m[sum];
for(int i=ss.length()-1;i>=0;i--)
printf("%c",ss[i]);
printf("\n");
}
}
return 0;
}
标签：1043,HDU,15,int,puzzle,tiles,键值,Eight,include 来源： https://blog.csdn.net/zab1998/article/details/96735179