hdu1043Eight(dbfs)
Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9152 Accepted Submission(s): 2466
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
South Central USA 1998 (Sepcial Judge Module By JGShining)
Recommend
JGShining
题目大意：经典题，不解释。
题目分析：杭电的测试数据比poj强，poj乱搞也能过，hdu卡单向bfs，双向比较保险，高级点的话就A*吧。
传说此题不做人生不完整。。。
详情请见代码：
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1000005;//poj4048K 0MS...
long fac[] = {1,1,2,6,24,120,720,5040,40320,362880};
int flag[2][363000];//记录前驱后继
char op[2][363000];//记录路径
bool ok;
struct node
{
int val,step,pos;
char state[9];
}ss,now;
int start,end;
struct que
{
struct node t[N];
int head,tail;
void init()
{
head = tail = 0;
}
bool empty()
{
return head == tail;
}
void push(struct node a)
{
t[tail] = a;
tail ++;
if(tail >= N)
tail -= N;
}
struct node top()
{
return t[head];
}
void pop()
{
head ++;
if(head >= N)
head -= N;
}
}q[2];
int contor()
{
int i,j;
int tmp,num;
num = 0;
for(i = 0;i < 9;i ++)
{
tmp = 0;
for(j = i + 1;j < 9;j ++)
if(ss.state[j] < ss.state[i])
tmp ++;
num += fac[9 - i - 1] * tmp;
}
return num;
}
void output(int cur)
{
if(cur == start)
return;
output(flag[0][cur]);
putchar(op[0][cur]);
}
void print()
{
int i,root;
output(ss.val);//从起点到相遇点的路径输出
root = ss.val;
do
{
switch(op[1][root])
{
case 'u':putchar('d');break;
case 'd':putchar('u');break;
case 'l':putchar('r');break;
case 'r':putchar('l');break;
}
root = flag[1][root];
}while(root != end);//从相遇点到终点的路径输出，注意反向
putchar(10);
}
void dbfs()
{
int i,j;
q[0].init();
q[1].init();
ss.step = 0;
q[0].push(ss);
ss.val = 0;
for(i = 0;i < 9;i ++)
ss.state[i] = '1' + i;
ss.pos = 8;
q[1].push(ss);
flag[1][end] = end;
flag[0][start] = start;
j = 0;
while(!q[0].empty() && !q[1].empty())
{
for(i = 0;i < 2;i ++)
{
while(!q[i].empty() && q[i].top().step == j)
{
now = q[i].top();
//printf("%d %d %d\n",i,now.step,now.val);
//system("pause");
q[i].pop();
if(now.pos > 2)//up
{
ss = now;
ss.step ++;
ss.state[ss.pos] ^= ss.state[ss.pos - 3] ^= ss.state[ss.pos] ^= ss.state[ss.pos - 3];
ss.pos -= 3;
ss.val = contor();
if(flag[i][ss.val] == -1)
{
flag[i][ss.val] = now.val;
op[i][ss.val] = 'u';
if(flag[1 - i][ss.val] != -1)
{
print();
return;
}
q[i].push(ss);
}
}
if(now.pos < 6)//down
{
ss = now;
ss.step ++;
ss.state[ss.pos] ^= ss.state[ss.pos + 3] ^= ss.state[ss.pos] ^= ss.state[ss.pos + 3];
ss.pos += 3;
ss.val = contor();
if(flag[i][ss.val] == -1)
{
flag[i][ss.val] = now.val;
op[i][ss.val] = 'd';
if(flag[1 - i][ss.val] != -1)
{
print();
return;
}
q[i].push(ss);
}
}
if(now.pos % 3)//left
{
ss = now;
ss.step ++;
ss.state[ss.pos] ^= ss.state[ss.pos - 1] ^= ss.state[ss.pos] ^= ss.state[ss.pos - 1];
ss.pos --;
ss.val = contor();
if(flag[i][ss.val] == -1)
{
flag[i][ss.val] = now.val;
op[i][ss.val] = 'l';
if(flag[1 - i][ss.val] != -1)
{
print();
return;
}
q[i].push(ss);
}
}
if(now.pos % 3 != 2)//right
{
ss = now;
ss.step ++;
ss.state[ss.pos] ^= ss.state[ss.pos + 1] ^= ss.state[ss.pos] ^= ss.state[ss.pos + 1];
ss.pos ++;
ss.val = contor();
if(flag[i][ss.val] == -1)
{
flag[i][ss.val] = now.val;
op[i][ss.val] = 'r';
if(flag[1 - i][ss.val] != -1)
{
print();
return;
}
q[i].push(ss);
}
}
}
}
j ++;
}
//printf("out loop\n");
printf("unsolvable\n");
}
int main()
{
char s[3];
while(scanf("%s",s) != EOF)
{
ss.state[0] = s[0];
if(ss.state[0] == 'x')
{
ss.state[0] = '9';
ss.pos = 0;
}
for(int i = 1;i < 9;i ++)
{
scanf("%s",s);
if(s[0] == 'x')
{
s[0] = '9';
ss.pos = i;
}
ss.state[i] = s[0];
}
ss.val = contor();
start = ss.val;
end = 0;
int so = 0;
for(int ii = 0;ii < 9;ii ++)//逆序数偶数才有解，不包括空格！！
{
if(ss.state[ii] == '9')
continue;
for(int jj = 0;jj < ii;jj ++)
{
if(ss.state[jj] == '9')
continue;
if(ss.state[jj] < ss.state[ii])
so ++;
}
}
//printf("%d\n",so);
if(so % 2)
{
printf("unsolvable\n");
continue;
}
memset(flag,-1,sizeof(flag));
dbfs();
}
return 0;
}
//hdu421MS 4092K
//poj4052K 16MS