/*
分析:
A*搜索的状态包含了布局和在搜索树中的深度.
因为在优先队列的优先级是f,所以无法实现布局的二分查找.但是在open和close表中又要查找布局.不可行..
通过hash表来记录已经搜索过的布局的最小f,那么不需要在堆中查找,如果f较大,不入队列,也能保证最短步数.
按照从上到下,从左到右的顺序,八数码就是123456780,9个数字的排列是9!.
*/#include <iostream>
#include <cmath>
#include <stack>
using namespace std;const int MAX = 1 << 20;
int step;
const int EndPos[9][2] = {{2, 2}, {0, 0}, {0, 1}, {0, 2}, {1, 0}, {1, 1}, {1, 2}, {2, 0}, {2, 1}};
const int hashFac[9] = {1, 2, 6, 24, 120, 720, 5040, 40320};
int hash_table[MAX];
int heap_last;//堆中最后一个元素的下标(可插入位置-1)
const int dir[4][2] = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}};//r,u,l,d
const int EndBoard[3][3] = {1, 2, 3, 4, 5, 6, 7, 8, 0};struct Status
{int board[3][3];int r, c;int d;//移动方向int g,h;Status *father;
}start, End, Queue[MAX], Step[MAX];;//用堆来实现的优先队列,第一个元素的小标是1int HFun(const Status &s)
{int tile, cnt = 0;for(int r = 0; r < 3; r++)for(int c = 0; c < 3; c++){tile = s.board[r][c];if(0 != tile)cnt += abs(EndPos[tile][0] - r) + abs(EndPos[tile][1] - c);}return cnt;
}
bool operator<(const Status &s1, const Status &s2)
{return ((s1.g + s1.h) < (s2.g + s2.h));//这里刚开始的时候把h也写成了g..
}int HashVal(const Status &s)//全排列hash函数,不会冲突
{int i,j,k = 0,array[9];for(i = 0; i < 3; i++)for(j = 0; j < 3; j++)array[k++] = s.board[i][j];int cnt = 0, res = 0;for(i = 0; i < 9; i++){cnt = 0;for(j = 0; j < i; j++)if(array[j] > array[i])cnt++;res += cnt * hashFac[i];}       return res;
}bool Empty()
{return 0 == heap_last;
}//向下维护小顶堆
void down_min_heap()
{Status s = Queue[1];int i,j;for(i = 1, j = i<<1; j <= heap_last; j = j << 1){if(j + 1 <= heap_last && Queue[j + 1] < Queue[j])//较小孩子节点的下标j++;if(Queue[j] < s){Queue[i] = Queue[j];//元素上移i = j;}elsebreak;}Queue[i] = s;
}Status Pop()
{Status s = Queue[1];Queue[1] = Queue[heap_last--];down_min_heap();return s;
}/*
向上维护小顶堆
父亲结点必定小于另外一个孩子结点
所以如果该结点小于父亲结点,那么也小于另外一个孩子结点,直接往上移动就好
否则已经满足小顶堆特性
*/
void up_min_heap()
{int i,j;Status s = Queue[heap_last];for(j = heap_last, i = j>>1; i >= 1; i = i >> 1){if(s < Queue[i]){Queue[j] = Queue[i];j = i;}elsebreak;}Queue[j] = s;
}void Push(const Status &s)
{Queue[++heap_last] = s;up_min_heap();
}void Trace();
bool Astar()
{memset(hash_table, 0, sizeof(hash_table));int r, c, nr, nc, tile, f, hashVal;start.g = 0;start.h = HFun(start);hash_table[HashVal(start)] = start.g + start.h;Push(start);Status s, ns;while(!Empty()){s = Pop();Step[step++] = s;if(!memcmp(s.board, EndBoard, sizeof(EndBoard))){End.father = &s;Trace();return true;}r = s.r; c = s.c;for(int d = 0; d < 4; d++){if(abs(s.d - d) == 2){continue;}nr = r + dir[d][0]; nc = c + dir[d][1];if(nr < 0 || nr > 2 || nc < 0 || nc > 2)continue;ns = s;ns.d = d;tile = ns.board[nr][nc];ns.board[r][c] = tile;ns.board[nr][nc] = 0;ns.r = nr; ns.c = nc;//这里刚开始写成了c,结果调试的时候发现经常2个0..无限蛋疼!!!ns.g++;ns.h = HFun(ns);f = ns.g + ns.h;hashVal = HashVal(ns);if(hash_table[hashVal] < f)//未访问或者估价更小的状态{hash_table[hashVal] = f;ns.father = &Step[step - 1];Push(ns);}}//cout<<heap_last<<endl;}return false;
}void input()
{int tile;for(int r = 0; r < 3; r++)for(int c = 0; c < 3; c++){cin>>tile;if(cin.fail()){cin.clear();cin.ignore(1, ' ');start.r = r;start.c = c;start.board[r][c] = 0;}else{start.board[r][c] = tile;}/*if(!tile){start.r = r;start.c = c;}*/}start.d = 100;//第一次不定义方向
}bool Solveable()
{int i, j, array[9], k = 0, sum = 0;for(i = 0; i < 3; i++)for(j = 0; j < 3; j++)array[k++] = start.board[i][j];for(i = 8; i >= 0; i--)for(j = i - 1; j >= 0; j--)if(array[j] && array[j] < array[i])sum++;if(sum % 2 == 0)return true;elsereturn false;
}void Trace()
{stack<int> trace;Status *p = End.father;while(p){trace.push(p->d);p = p->father;}while(!trace.empty()){int d = trace.top();trace.pop();switch(d){case 0:cout<<'r';break;case 1:cout<<'u';break;case 2:cout<<'l';break;case 3:cout<<'d';break;}}cout<<endl;
}int main()
{//freopen("in.txt", "r", stdin);input();if(Solveable())Astar();elsecout<<"unsolvable"<<endl;return 0;
}