POJ1077&HDU1043 Eight 八数码第八境界 IDA* hash 康托展开 奇偶剪枝

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

终于把八数码修炼第八境界了，先为自己鼓个爪！

第八境界用到了IDA*，hash，康托展开和逆展开，哈曼顿距离做下界剪枝，奇偶剪枝

不过其实还没有第七境界快，但是第八境界大大节约了内存，而且实现起来比A*容易多了。

#include <cstdio>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <algorithm>
#include <map>
#include <stack>
using namespace std;
const int MAXN = 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 + 10;
const int INF = 0x3f3f3f3f;

struct Path {
int dir;
int next;
Path(int dir = 0, int next = 0) {
this->dir = dir;
this->next = next;
}
} path[MAXN];

int jie[10] = {1, 1};
int dirx[] = {-1, 0, 1, 0};
int diry[] = {0, -1, 0, 1};
char dir[] = "uldr";
int minAns;

string Tidy(char* s) {
int len = strlen(s);
string ans = "";
for (int i = 0; i < len; ++i) {
if (s[i] == 'x') s[i] = '9';
if (s[i] != ' ') ans += s[i];
}
return ans;
}

int IndexOfX(string s) {
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '9') return i;
}
return -1;
}

bool InRange(int x, int y) {
return x >= 0 && x < 3 && y >= 0 && y < 3;
}

int Hash(string s) {
int ans = 0;
for (int i = 0; i < 9; ++i) {
int rev = 0;
for (int j = i + 1; j < 9; ++j) {
if (s[i] > s[j]) ++rev;
}
ans += rev * jie[8 - i];
}
return ans;
}

string RevHash(int val) {
string ans = "";
bool tag[10] = {};
for (int i = 0; i < 9; ++i) {
int tNum = val / jie[8 - i] + 1;
for (int j = 1; j <= tNum; ++j) {
if (tag[j]) ++tNum;
}
val %= jie[8 - i];
ans += tNum + '0';
tag[tNum] = true;
}
return ans;
}

int H(string s) {
int h = 0;
int index = 0;
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
int num = s[index++] - '1';
if (num == 8) continue;
int dx = abs(num / 3 - i);
int dy = abs(num % 3 - j);
h += dx + dy;
}
}
return h;
}

void PutPath(int i) {
while (path[i].next != -1) {
putchar(dir[path[i].dir]);
i = path[i].next;
}
putchar('\n');
}

int DFS(int node, int index, int dir, int g) {
if (node == 0) return g;

string s = RevHash(node);
if (g + H(s) > minAns) return INF;

for (int i = 0; i < 4; ++i) {
if (abs(i - dir) == 2) continue;
int nx = index / 3 + dirx[i];
int ny = index % 3 + diry[i];
if (!InRange(nx, ny)) continue;

int nIndex = nx * 3 + ny;
swap(s[index], s[nIndex]);
int nNode = Hash(s);

int temp = DFS(nNode, nIndex, i, g + 1);
if (temp != INF) {
path[node] = Path(i, nNode);
return temp;
}
swap(s[index], s[nIndex]);
}
return INF;
}

void IDAStar(string s) {
path[0] = Path(0, -1);
minAns = H(s);
while (DFS(Hash(s), IndexOfX(s), 0, 0) == INF) ++minAns;
PutPath(Hash(s));
}

int main() {
#ifdef NIGHT_13
freopen("in.txt", "r", stdin);
#endif
for (int i = 1; i < 10; ++i) {
jie[i] = jie[i - 1] * i;
}

char s[100];
while (gets_s(s) != NULL) {
string ma = Tidy(s);
int rev = 0;

for (int i = 0; i < 9; ++i) {
if (ma[i] == '9') continue;
for (int j = i + 1; j < 9; ++j) {
if (ma[i] > ma[j]) ++rev;
}
}
if (rev & 1) puts("unsolvable");
else IDAStar(ma);
}
return 0;
}