F - Eight Puzzle
Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
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 a4 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. To simplify this problem, you should print the minimum steps only.
Input
There are multiple test cases.
For each test case, 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.Otherwise, output an integer which equals the minimum steps.
Sample input and output
| Sample Input | Sample Output |
|---|---|
1 2 x 4 5 3 7 8 6 |
2 |
Hint
Any violent algorithm may gain TLE. So a smart method is expected.
The data used in this problem is unofficial data prepared by hzhua. So any mistake here does not imply mistake in the offcial judge data.
解题报告:
HINT部分已经知道本题数据很强，因此暂时不考虑普通bfs,那么我们可以考虑双广和A*两种算法..
关于两种方法的结果:
1.A*超时
2.双广AC
可能的原因：首先如果没有解，都退化成普通bfs，这点并没有区别，那么只能说明在有解的时候双广比A*高效很多
当然效率还可以进一步提升，那就是奇偶性剪枝，想了解的话可以百度
双广代码(有奇偶性剪枝)
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std; const int maxhashsize = + ;
const int maxstatussize = 1e6 + ;
int vis1[maxhashsize],vis2[maxhashsize];
int fac[];
int dir[][] = {-,,,,,-,,}; typedef struct status
{
char s[] , step;
int val;
}; status q1[maxstatussize],q2[maxstatussize]; int gethashvalue(const status &x)
{
int res = ;
for(int i = ; i < ; ++ i)
{
int cot = ;
for(int j = i+ ; j < ; ++ j)
if (x.s[i] > x.s[j])
cot++;
res += fac[-i]*cot;
}
return res;
} status st,ed; int bfs()
{
int front1 = , rear1 = ;
int front2 = , rear2 = ;
q1[rear1++] = st;
q2[rear2++] = ed;
if (st.val == ed.val )
return ;
while(front1 < rear1 || front2 < rear2)
{
//
{
status ns = q1[front1++];
int x,y,step = ns.step,oripos;
for(int i = ; i < ; ++ i)
if (!ns.s[i])
{
x = i / ;
y = i % ;
oripos = i;
break;
}
for(int i = ; i < ; ++ i)
{
int newx = x + dir[i][];
int newy = y + dir[i][];
if (newx >= || newx < || newy >= || newy < )
continue;
int newpos = newx*+newy;
status ss;
memcpy(&ss,&ns,sizeof(struct status));
swap(ss.s[newpos],ss.s[oripos]);
int newhash = gethashvalue(ss);
if (vis1[newhash] != -)
continue;
ss.step ++ ;
if (vis2[newhash] != -)
return ss.step + vis2[newhash];
vis1[newhash] = ss.step;
ss.val = newhash;
q1[rear1++] = ss;
}
}
//**************************
{
status ns = q2[front2++];
int x,y,step = ns.step,oripos;
for(int i = ; i < ; ++ i)
if (!ns.s[i])
{
x = i / ;
y = i % ;
oripos = i;
break;
}
for(int i = ; i < ; ++ i)
{
int newx = x + dir[i][];
int newy = y + dir[i][];
if (newx >= || newx < || newy >= || newy < )
continue;
int newpos = newx*+newy;
status ss;
memcpy(&ss,&ns,sizeof(struct status));
swap(ss.s[newpos],ss.s[oripos]);
int newhash = gethashvalue(ss);
if (vis2[newhash] != -)
continue;
ss.step ++ ;
if (vis1[newhash] != -)
return ss.step + vis1[newhash];
vis2[newhash] = ss.step;
ss.val = newhash;
q2[rear2++] = ss;
}
}
}
return -;
} bool input()
{
char ch = getchar();
if (ch == EOF) return false;
memset(vis1,-,sizeof(vis1));
memset(vis2,-,sizeof(vis2));
if (ch == 'x')
st.s[] = ;
else
st.s[] = ch-'';
getchar();
for(int i = ; i <= ; ++ i)
{
ch = getchar();getchar();
if (ch == 'x')
st.s[i] = ;
else
st.s[i] = ch-'';
}
st.step = ;
vis1[gethashvalue(st)] = ; // Init for vis
st.val = gethashvalue(st);
vis2[gethashvalue(ed)] = ;
ed.val = gethashvalue(ed);
return true;
} int main(int argc,char *argv[])
{
fac[] = ;
for(int i = ; i <= ; ++ i)
fac[i] = i*fac[i-];
for(int i = ; i < ; ++ i)
ed.s[i] = i + ;
ed.s[] = ;
ed.step = ;
while(input())
{
int sum = ;
//奇偶性判断
for(int i = ; i < ; ++ i)
{
if (st.s[i] == )
continue;
for(int j = ; j < i ; ++ j)
if (st.s[j] > st.s[i])
sum++;
}
if ( sum % & )
{
cout << "unsolvable" << endl;
continue;
}
int ans = bfs();
if (ans == -)
cout << "unsolvable" << endl;
else
cout << ans << endl;
}
return ;
}