HDU 1043 Eight (A* + HASH + 康托展开)

Eight

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13956 Accepted Submission(s): 3957 Special Judge

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2 3 4 1 5 x 7 6 8

[align=left]Sample Output[/align]

ullddrurdllurdruldr

这题活生生搞了五天，为此学了A*和康托展开，虽然自己写出来了但还不是非常明白，有句话说“如果你真的懂了那么你就应该自己动手做一个”，看来我还没真的懂。
首先把每次矩阵的格局存结构里，然后对整个矩阵用康托展开来进行哈希，得到一个值，以此来判断此种情况是否出现过或是是否到达了终点，有一个很好的剪枝就是用奇偶逆序数来判断是否可解，因为最终情况的逆序数是0，而每次变换不会改变逆序数的奇偶，所以如果初始情况的逆序数是奇数的话就不可解。
不明白的地方是A*的估值函数为何不用f = g + h，而是要用h和g分别作为两个参数来比较，刚开始的时候我把两个参数的位置弄反了， T了一天。都是泪。

#include <iostream>
#include <cmath>
#include <string>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using    namespace    std;

const    int    SIZE = 5;
const    int    GOAL = 46233;
const    int    HASH[] = {40320,5040,720,120,24,6,2,1,1};
const    int    UP_DATE[][2] = {{0,-1},{0,1},{-1,0},{1,0}};
int    PATH[600000];
int    PRE[600000];
struct    Node
{
int    map[SIZE][SIZE];
int    x,y;
int    h,g;
int    hash;
bool    operator <(const Node a) const
{
return    h != a.h ? h > a.h : g > a.g;
}
};

bool    solve_able(const Node & r);
bool    check(const int,const int);
void    cal_hash(Node & r);
void    cal_h(Node & r);
void    search(const Node & r);
void    show(void);
int    main(void)
{
Node    first;
char    s[50];

while(gets(s))
{
int    k = 0;
memset(PRE,-1,sizeof(PRE));
memset(PATH,-1,sizeof(PATH));
for(int i = 1;i <= 3;i ++)
for(int j = 1;j <= 3;j ++)
{
if(s[k] >= '1' && s[k] <= '9')
first.map[i][j] = s[k] - '0';
else    if(s[k] == 'x')
{
first.map[i][j] = 0;
first.x = i;
first.y = j;
}
else
j --;
k ++;
}
if(!solve_able(first))
{
printf("unsolvable\n");
continue;
}
cal_hash(first);
if(first.hash == GOAL)
{
puts("");
continue;
}
PATH[first.hash] = -2;
first.g = 0;
cal_h(first);
search(first);
}

return    0;
}

bool    solve_able(const Node & r)
{
int    sum = 0,count = 0;
int    temp[10];

for(int i = 1;i <= 3;i ++)
for(int j = 1;j <= 3;j ++)
{
temp[count] = r.map[i][j];
count ++;
}
for(int i = 0;i < 9;i ++)
for(int j = i + 1;j < 9;j ++)
if(temp[j] < temp[i] && temp[j] && temp[i])
sum ++;
return    !(sum & 1);
}

bool    check(const int    x,const    int y)
{
if(x >= 1 && x <= 3 && y >= 1 && y <= 3)
return    true;
return    false;
}

void    cal_hash(Node & r)
{
int    sum = 0,count = 0,box;
int    temp[10];

for(int i = 1;i <= 3;i ++)
for(int j = 1;j <= 3;j ++)
{
temp[count] = r.map[i][j];
count ++;
}
for(int i = 0;i < 9;i ++)
{
box = 0;
for(int j = i + 1;j < 9;j ++)
if(temp[j] < temp[i])
box ++;
sum += (box * HASH[i]);
}
r.hash = sum;
}

void    search(Node const & r)
{
Node    cur,next;

priority_queue<Node>    que;
que.push(r);
while(!que.empty())
{
cur = que.top();
que.pop();
for(int i = 0;i < 4;i ++)
{
next = cur;
next.x = cur.x + UP_DATE[i][0];
next.y = cur.y + UP_DATE[i][1];
if(!check(next.x,next.y))
continue;
swap(next.map[cur.x][cur.y],next.map[next.x][next.y]);
cal_hash(next);

if(PATH[next.hash] == -1)
{
PATH[next.hash] = i;
PRE[next.hash] = cur.hash;
next.g ++;
cal_h(next);
que.push(next);
}
if(next.hash == GOAL)
{
show();
return    ;
}
}
}

}

void    cal_h(Node & r)
{
int    ans = 0;
for(int i = 1;i <= 3;i ++)
for(int j = 1;j <= 3;j ++)
if(r.map[i][j])
ans += abs(i - ((r.map[i][j] - 1) / 3 + 1)) + abs(j - ((r.map[i][j] - 1) % 3 + 1));
r.h = ans;
}

void    show(void)
{
string    ans;
int    hash = GOAL;

ans.clear();
while(PRE[hash] != -1)
{
switch(PATH[hash])
{
case    0:ans += 'l';break;
case    1:ans += 'r';break;
case    2:ans += 'u';break;
case    3:ans += 'd';break;
}
hash = PRE[hash];
}
for(int i = ans.size() - 1;i >= 0;i --)
printf("%c",ans[i]);
cout << endl;
}