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
1 2 3 x 4 6 7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
Output
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
終於把八數碼修煉第八境界了,先爲自己鼓個爪!
第八境界用到了IDA*,hash,康託展開和逆展開,哈曼頓距離做下界剪枝,奇偶剪枝
不過其實還沒有第七境界快,但是第八境界大大節約了內存,而且實現起來比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;
}