Eight
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 37738   Accepted: 15932   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:

 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

 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

 
年轻人的第一道A* 答案可能多解
我的这个程序输入样例输入得到的输出是 ldruullddrurdllurrd
跟样例输出不一样 搞的我郁闷了半天
 
注意方向的选取 还有位置的选择
它是
1 2 3
4 5 6
7 8 9
 
康拓展开
类似哈希 给定一个排列组合的数  如123456789 怎么去判定有没有遍历过这个数 可以用康拓展开 八数码问题有9位 直接开1e9的数组可能爆内存 并且有很多数字是用不到的 因为每个数字只出现一次 可能会有123456789 但是绝对不会出现123456788 以此类推 这样浪费了很多空间
由此引出康拓展开 由这个数是第几小的数来作表示 对N个不同的数的排列组合 一共有N!种排列组合 他们的大小都不一样 所以只需要N!的空间就足够存放所有的状态
要注意自己定义的康拓的范围
其中 递增顺序排列是最小的数 即123456789  987654321是最大的数
我这里写的是goal = 1,  所以 cantor是返回v+1避免出现0, 要是 goal = 0, 则cantor返回v就行了 以此类推
记忆方法: 只要记住顺序排列是最小的 它的v是最小的 在以下代码片中要让return v+1 等于1 容易知道下划线部分应该填  ar[j] < ar[i]
int contor(int ar[]) {
int v = ;
for (int i = ; i < ; i++) {
int num = ;
for (int j = i + ; j < ; j++) {
if (_______) num++;
}
v += num * fac[ - i];
}
return v + 1;
}
 
还有个注意点是曼哈顿距离不算x点
短即是正义 我短我骄傲
 #include <stdio.h>
#include <queue>
#include <algorithm>
using namespace std;
int fac[];
int dx[] = {-, , , };
int dy[] = {, , , -};
char d[] = {'u','d','r','l'};
const int goal = , maxv = 4e5;
bool vis[maxv];
int pre[maxv];
char dir[maxv];
int dish(int ar[]) {
int dis = ;
for (int i = ; i < ; i++) {
if (ar[i] == ) continue;
int v = ar[i] - ;
int rx = i / , ry = i % ;
int x = v / , y = v % ;
dis += abs(x - rx) + abs(ry - y);
}
return dis;
}
int contor(int ar[]) {
int v = ;
for (int i = ; i < ; i++) {
int num = ;
for (int j = i + ; j < ; j++) if (ar[j] < ar[i]) num++; v += num * fac[ - i];
}
return v + ;
}
struct node {
int g, h, f, sta, xpos;
int s[];
void init() {
g++; h=dish(s); f=h+g; sta=contor(s);
}
bool operator < (const node x)const {
return f > x.f;
}
};
priority_queue<node> q;
bool Astar(node now) {
fill(vis, vis + maxv, );
while (!q.empty()) q.pop();
now.g = -;
now.init();
q.push(now);
pre[now.sta] = -;
vis[now.sta] = ;
while (!q.empty()){
now = q.top(); q.pop();
if (now.sta == goal) return true;
int xpos = now.xpos;
int x = xpos / , y = xpos % ;
for (int i = ; i < ; i++) {
int tx = x + dx[i];
int ty = y + dy[i];
if (tx < || ty < || tx >= || ty >= ) continue;
int newp = tx * + ty;
node tp = now;
tp.xpos = newp;
swap(tp.s[newp], tp.s[xpos]);
tp.init(); if (vis[tp.sta]) continue;
vis[tp.sta] = ;
pre[tp.sta] = now.sta;
dir[tp.sta] = d[i];
q.push(tp);
}
}
return false;
}
void Print(int sta) {
if (pre[sta] == -) return;
Print(pre[sta]);
printf("%c", dir[sta]);
}
int main() {
fac[] = ;
for (int i = ; i < ; i++) fac[i] = i * fac[i - ];
char str[];
while (gets(str)) {
int cnt = ;
node now;
for (int i = ; str[i]; i++) {
if (str[i] == ' ') continue;
if (str[i] == 'x') {
now.xpos = cnt;
now.s[cnt++] = ;
}
else
now.s[cnt++] = str[i] - '';
}
if (Astar(now)) Print(goal);
else printf("unsolvable");
printf("\n");
}
return ;
}
 

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