POJ 1077 Eight

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

Source

South Central USA 1998

解题思路：利用BFS与康托展开状态判重进行最优解的寻找，同时记录路径。

AC代码：

#include <stdio.h>
#include <string.h>
struct node {
	int op;
	int state[9];
	int parent;
	int dis;
};
struct node queue[362880];
//一些阶乘常数
long long factory[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
int vis[362880];
int dir[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}}; // r l u d
int operation[362880];
int start[9], goal[9] = {1, 2, 3, 4, 5, 6, 7, 8, 0};
int Cantor(int str[], int n) {
	long long res = 0;
	for (int i = 0; i < n; i++) {
		int counted = 0;
		for (int j = i + 1; j < n; j++) {
			if (str[i] > str[j])
				++counted;
		}
		res += counted * factory[n - i - 1];
	}
	if (!vis[res]) {
		vis[res] = 1;
		return 1;
	} else {
		return 0;
	}
}
void print_way(int n) {
	int top = 0;
	operation[top] = n;
	while (queue[operation[top]].parent != -1) {
		top++;
		operation[top] = queue[operation[top - 1]].parent;
	}
	for (int i = top; i >= 0; i--) {
		switch (queue[operation[i]].op) {
			case 0: {
				printf("r");
			}
			break;
			case 1: {
				printf("l");
			}
			break;
			case 2: {
				printf("u");
			}
			break;
			case 3: {
				printf("d");
			}
			break;
		}
	}
}
int bfs() {
	int front = 0, rear = 1;
	memcpy(queue[front].state, start, sizeof(queue[front].state));
	queue[front].dis = 0;
	queue[front].op = -1;
	queue[front].parent = -1;
	int x, y, z;
	while (rear - front) {

		struct node head  = queue[front];
		for (int i = 0; i < 9; i++) {
			if (head.state[i] == 0) z = i;
		}
		x = z % 3;
		y = z / 3;
		int newx, newy, newz;
			if (!memcmp(head.state, goal, sizeof(goal))) {				
							print_way(front);
							return 0;
						}

		for (int i = 0; i < 4; i++) {
			newx = x + dir[i][0];
			newy = y + dir[i][1];
			newz = newy * 3 + newx;
			if (newx >= 0 && newx < 3 && newy >= 0 && newy < 3) {
				int temp_;
				struct node temp;
				memcpy(temp.state, head.state, sizeof(temp.state));
				temp_ = temp.state[z];
				temp.state[z] = temp.state[newz];
				temp.state[newz] = temp_;
				temp.dis = head.dis + 1;
				temp.op = i;
				temp.parent = front;
				if (Cantor(temp.state,9)) {
					queue[rear++] = temp;
				}
			
			}
		}
		front++;
	}
	printf("unsolvable\n");
	return -1;
}
int main() {
	char ch;
	for (int i = 0; i < 9; i++) {
		scanf("%c", &ch);
		if (ch == ' ') {
			i--;
			continue;
		}
		if (ch == 'x') start[i] = 0;
		else if (ch != ' ') start[i] = ch - '0';
	}
	bfs();
}