原题链接


Eight


Time Limit: 1000MS


Memory Limit: 65536K

Total Submissions: 29854


Accepted: 12989


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


1.广搜,打表

从终点反向广搜,记录能够到达的所有点,打表.(hdu上109ms)

#include <cmath>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
#include <queue>
#define maxn 400005
using namespace std;
  
struct Node{
	char p[10];
	int x, y, f;
	char ch;
}node[maxn];
int vis[maxn], d[10];
int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
int k[4] = {'u', 'd', 'l', 'r'}, m = 1;
int cal(char *p){
	int ans = 0;
	for(int i = 0; i < 9; i++){
		int h = 0;
		for(int j = i+1; j < 9; j++)
		 if(p[i] > p[j])
		  h++;
		ans += h * d[8-i]; 
	}
	return ans;
}
void Bfs(){
	
	char t[10] = "123456780";
	vis[cal(t)] = 1;
	strcpy(node[1].p, t);
	node[1].x = 2, node[1].y = 2;
	node[1].f = -1;
	queue<int> q;
	q.push(1);
	while(!q.empty()){
		int h = q.front();
		q.pop();
		for(int i = 0; i < 4; i++){
			int x = node[h].x + dir[i][0];
			int y = node[h].y + dir[i][1];
			if(x >= 0 && x < 3 && y >= 0 && y < 3){
				strcpy(t, node[h].p);
				swap(t[node[h].x*3+node[h].y], t[x*3+y]);
				int e = cal(t);
				if(vis[e] == 0){
					++m;
					vis[e] = m;
					strcpy(node[m].p, t);
					node[m].x = x;
					node[m].y = y;
					node[m].ch = k[i];
					node[m].f = h;
					q.push(m);
				}
			}
		}
	}
}
void print(int e){
	
	vector<char> v;
	while(node[e].f != -1){
		v.push_back(node[e].ch);
		e = node[e].f;
	}
	for(int i = 0; i < v.size(); i++)
	 putchar(v[i]);
	puts("");
}
int main(){
	
//	freopen("in.txt", "r", stdin);
	d[1] = 1;
	for(int i = 2; i <= 9; i++)
	 d[i] = d[i-1] * i; 
	Bfs();
	char g[50];
	while(gets(g)){
		
		int e = 0;
		for(int i = 0; g[i]; i++){
			if(g[i] != ' ')
			 g[e++] = g[i];
			if(g[i] == 'x')
			 g[e-1] = '0';
		}
		e = cal(g);
		if(vis[e])
			print(vis[e]);
		else
		    puts("unsolvable");
	}
	return 0;
}





2.双向广度优先搜索(hdu上904ms)

#include <iostream>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <queue>  
#define maxn 400005  
using namespace std;  
  
struct Node{  
    char p[10];  
    char ch;  
    int f, x, y;  
}node[maxn];  
int vis[maxn];  
int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};  
char k[2][4] = {{'r', 'l', 'd', 'u'}, {'l', 'r', 'u', 'd'}};  
int d[10], m;  
queue<int> q[2];
int cal(const char *p){  
      
    int ans = 0;  
    for(int i = 0; i < 9; i++){  
        int cnt = 0;  
        for(int j = i+1; j < 9; j++){  
            if(p[i] > p[j])  
             cnt++;  
        }  
        ans += d[8-i] * cnt;  
    }  
    return ans;  
}  
void Init(){
	
	d[1] = 1;
	for(int i = 2; i < 9; i++)
	 d[i] = d[i-1] * i;
	while(!q[0].empty())
	 q[0].pop();
	while(!q[1].empty())
	 q[1].pop();
	memset(vis, 0, sizeof(vis));
	const char *p = "123456780";
	strcpy(node[2].p, p);
	node[2].x = 2, node[2].y = 2, node[2].f = -1;
	vis[cal(p)] = -2;
}  
bool judge(char *p){
	
	int cnt = 0;
	for(int i = 0; i < 9; i++)
	 for(int j = i+1; j < 9; j++){	
	 	if(p[i] != '0' && p[j] != '0' && p[i] > p[j])
	 	  cnt++;
	 }
	 if(cnt&1)return false;
	 return true;
}
void print1(int e){  
      
    if(node[e].f != -1){
    	print1(node[e].f);
    	putchar(node[e].ch);
    }
}  
void print2(int e){  
        
    while(node[e].f != -1){  
        putchar(node[e].ch);  
        e = node[e].f;  
    }  
    puts("");
}  
bool BFS(queue<int> &q, char *kk, int u){  
      
    char p[10];
    int t = q.front();
	q.pop();  
    for(int i = 0; i < 4; i++){  
        int x = node[t].x + dir[i][0];  
        int y = node[t].y + dir[i][1];
		if(x >= 0 && x < 3 && y >= 0 && y < 3){
			strcpy(p, node[t].p);
			swap(p[node[t].x*3+node[t].y], p[x*3+y]); 
			int u1 = cal(p);
			if(vis[u1] == 0){
				++m;
				node[m].x = x, node[m].y = y;
				node[m].f = t, node[m].ch = kk[i];
				strcpy(node[m].p, p);
				vis[u1] = m * u;
				q.push(m);
			}
			else if(vis[u1] * u < 0){
				if(u == 1){
					print1(t);
					putchar(kk[i]);
					print2(-vis[u1]);
				}
				else{
					print1(vis[u1]);
					putchar(kk[i]);
					print2(t);
				}
				return true;
			}
		}  
    }
    return false;
}  
int main(){  
      
  // freopen("in.txt", "r", stdin);  
    char g[50];  
    while(gets(g)){  
        Init();
        int v = -1;  
        for(int i = 0; g[i]; i++){  
            if(g[i] != ' ')  
              node[1].p[++v] = g[i];  
            if(g[i] == 'x'){
             node[1].p[v] = '0';
             node[1].x = v / 3;
             node[1].y = v % 3;
		    }
        }  
        if(!judge(node[1].p)){
        	puts("unsolvable");
        	continue;
        }
        node[1].p[v+1] = 0;
        node[1].f = -1;  
        vis[cal(node[1].p)] = 1;  
        q[0].push(1);  
        q[1].push(2);  
        int sign = 0;  
        m = 2;  
        while(!q[0].empty() && !q[1].empty()){
		    int e = 1;
			if(q[0].size() < q[1].size())
			 e = 0;  
            if(BFS(q[e], k[e], e == 0 ? 1 : -1)){  
                sign = 1;  
                break;  
            }  
        }  
        if(sign == 0){  
            puts("unsolvable");  
        }  
   }  
   return 0;  
}



3.启发式搜索(HDU717ms)



发中的估价是用估价函数表示的,如:f(n) = g(n) + h(n)


其中f(n) 是节点n的估价函数,g(n)是在状态空间中从初始节点到n节点的实际代价,h(n)是从n到目标节点最佳路径的估计代价。在这里主要是h(n)体现了搜索的启发信息,


#include <iostream>  
#include <algorithm>  
#include <queue>
#include <cstdio>
#include <cstring>
#include <vector>  
#include <map>
#define maxn 400005
using namespace std;  
typedef long long ll;

struct Node{
	char p[10];
	int pre;
	int w, h, x, y;
	char ch;
}node[maxn];
struct cmp{
	bool operator ()(const int &a, const int &b){
		int f1 = node[a].w + node[a].h;
		int f2 = node[b].w + node[b].h;
		return (f1 == f2 && node[a].h > node[b].h) || f1 > f2;
	}
};
int vis[maxn], d[10], e = 46233, m;
int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
char kk[4] = {'r', 'l', 'd', 'u'};
int cal(const char *p){	
	int ans = 0;
	for(int i = 0; i < 9; i++){
		int k = 0;
		for(int j = i + 1; j < 9; j++)
		 if(p[j] < p[i])
		  k++;
		ans += d[8-i] * k;
	}
	return ans;
}
bool judge(char *p){
	
	int cnt = 0;
	for(int i = 0; i < 9; i++)
	 for(int j = i+1; j < 9; j++)
	  if(p[i] != '0' && p[j] != '0' && p[i] > p[j])
	   cnt++;
	 if(cnt&1)
	  return false;
	 return true; 
}
int get_h(char *p){
	
	int ans = 0;
	for(int i = 0; i < 9; i++){
		if(p[i] != '0')
		ans += abs(i/3-(p[i]-'0'-1)/3) + abs(i%3-(p[i]-'0'-1)%3);
	}
	return ans;
}
void print(int v){
	if(node[v].pre != -1){
		print(node[v].pre);
		putchar(node[v].ch);
	}
}
bool  Bfs(){
	char p[10];
	priority_queue<int, vector<int>, cmp> q;
	q.push(1);
	int c = cal(node[1].p);
	if(c == e){
		puts("");
		return true; 
	}
	vis[c] = 1;
	while(!q.empty()){
		c = q.top();
		q.pop();
		for(int i = 0; i < 4; i++){
			int x = node[c].x + dir[i][0];
			int y = node[c].y + dir[i][1];
			if(x >= 0 && x < 3 && y >= 0 && y < 3){
			   strcpy(p, node[c].p);
			   swap(p[3*node[c].x+node[c].y], p[3*x+y]);
			   int v = cal(p);
			   if(vis[v] == 0){
			   	vis[v] = 1;
			   	++m;
			   	strcpy(node[m].p, p);
			   	node[m].w = node[c].w + 1;
			   	node[m].h = get_h(node[m].p);
			   	node[m].x = x, node[m].y = y;
			   	node[m].pre = c, node[m].ch = kk[i];
			   	q.push(m);
			   }
			   if(v == e){
			   	print(m);
			   	puts("");
			   	return true;
			   }	
			   
			}
		}
	
	}
	return false;
}
int main(){
	
//	freopen("in.txt", "r", stdin); 
	d[1] = 1;
	for(int i = 2; i <= 9; i++)
	  d[i] = d[i-1] * i;
	char g[50];
	while(gets(g)){
		memset(vis, 0, sizeof(vis));
		m = 1;
		int k = -1;
		for(int i = 0; g[i]; i++){
		  if(g[i] != ' ')
		     node[1].p[++k] = g[i];
		  if(g[i] == 'x'){
		     node[1].p[k] = '0';
		     node[1].x = k / 3;
			 node[1].y = k % 3; 
		  }
	    }
	    if(!judge(node[1].p)){
	    	puts("unsolvable");
	    	continue;
	    }
	    node[1].p[k+1] = 0;
	    node[1].pre = -1;
	    node[1].w = 0, node[1].h = get_h(node[1].p);	
		if(!Bfs())
		 puts("unsolvable");
	}   
	return 0;
}