[POJ 1077] Eight【双向BFS】

  • 2017-12-31
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Problem:

Time Limit: 1000MS Memory Limit: 65536K Special Judge

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 

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

Solution:

本来是 BFS 大水题,练了一下双向 BFS,处理输出时似乎复杂不少。

采用双向 BFS + 康托展开哈希

注意事项:

  1. #define 宏函数时一定要注意多打括号,避免运算时因优先级出错。
  2. 双向 BFS 处理相遇时由于要输出路径,必须分两类讨论。

康托展开(Cantor Expansion)

  • 将序列 {a1, a2, ..., an} 展开为正整数 X, 表示该序列为全排列中的第 X 项。
  • 设 ri 表示 ai+1 ~ an 中比 ai 小的个数,则
  • X = r1·(n - 1)! + r2·(n - 2)! + ... + ri·(n - i)! + ... + rn·0!

Code: O(n·n!), n为总格数 [3660K, 32MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
const int fac[9] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};
const int dx[4] = {0, 0,-1, 1};
const int dy[4] = {1,-1, 0, 0};

inline void getTile(int &t){
	char ch = getchar();
	while(ch < '0') ch = getchar();
	t = (ch == 'x') ? 0 : ch - '0'; 
}

struct Puzzle{
	int con[3][3];
	
	inline int getCantor(){  // Cantor Expansion Hash 
		int cantor = 0;
		for(register int i = 0; i < 9; i++){
			int cnt = 0;
			for(register int j = i + 1; j < 9; j++)
				if(con[i / 3][i % 3] > con[j / 3][j % 3]) cnt++;
			cantor += cnt * fac[8 - i];
		}
		return cantor;
	}
} s, t;

struct Queue{
	#define MAXNODE 362885
	
	Puzzle node[MAXNODE];
	int fr, re;
	
	inline void clear() {fr = re = 0;}
	
	inline bool empty() {return fr == re;}
	
	inline void push(const Puzzle &x) {node[re++] = x;}
	
	inline void pop() {if(!empty()) fr++;}
	
	inline Puzzle front() {return node[fr];}
} q;

int prev[MAXNODE], dir[MAXNODE];  // Positions of the queue serve as subscriptions
int vis[MAXNODE], qpos[MAXNODE];  // Cantor values serve as subscriptions

#define showDir(d) putchar((d) == 0 ? 'r' : ((d) == 1 ? 'l' : ((d) == 2 ? 'u' : 'd')))
// Beware of the priority of ^(XOR) used below, remember to add a bracket around each "d" !!!

inline void displayFormer(int pos){
	if(dir[pos] == -1) return;
	displayFormer(prev[pos]);
	showDir(dir[pos]);
}

inline void bidirectional_BFS(){
	q.clear();
	int sCantor = s.getCantor(), tCantor = t.getCantor();
	prev[0] = dir[0] = -1, vis[sCantor] = 1, qpos[sCantor] = 0, q.push(s);
	prev[1] = dir[1] = -1, vis[tCantor] = 2, qpos[tCantor] = 1, q.push(t);
	while(!q.empty()){
		Puzzle u = q.front();
		int uCantor = u.getCantor(), Bx, By;
		for(register int i = 0; i < 9; i++)
			if(!u.con[i / 3][i % 3]){
				Bx = i / 3, By = i % 3;
				break;
			}
		for(register int i = 0; i < 4; i++){
			int Nx = Bx + dx[i], Ny = By + dy[i];
			if(Nx < 0 || Nx >= 3 || Ny < 0 || Ny >= 3) continue;
			Puzzle v = u; swap(v.con[Bx][By], v.con[Nx][Ny]);  // Construct the new state
			int vCantor = v.getCantor();
			if(!vis[vCantor]){
				prev[q.re] = q.fr, dir[q.re] = (vis[uCantor] == 1 ? i : i ^ 1);
				vis[vCantor] = vis[uCantor], qpos[vCantor] = q.re;
				q.push(v);
			}  // Push the new-reached state at the rear of queue
			if(vis[vCantor] != vis[uCantor]){
				// Beware of the putting-together of 3 parts of the answer !!!
				if(vis[uCantor] == 1){  // When a state "u" transfered from origin meets with another state "v" already transfered from destination
					displayFormer(q.fr);  // Part 1: origin - u
					showDir(i);  // Part2: u - v
					for(register int curPos = qpos[vCantor]; dir[curPos] != -1; curPos = prev[curPos])
						showDir(dir[curPos]);  // Part 3: v - destination
				}
				else{  // When a state "u" transfered from destination meets with another state "v" already transfered from origin
					// All directions reverse
					displayFormer(qpos[vCantor]);  // Part 1: origin - v
					showDir(i ^ 1);  // Part 2: v - u
					for(register int curPos = q.fr; dir[curPos] != -1; curPos = prev[curPos])
						showDir(dir[curPos]);  // Part 3: u - destination
				}
				puts("");
				return;
			}
		}
		q.pop();
	}
	puts("unsolvable");
}

int main(){
	for(register int i = 0; i < 9; i++) getTile(s.con[i / 3][i % 3]);
	for(register int i = 0; i < 8; i++) t.con[i / 3][i % 3] = i + 1;
	t.con[2][2] = 0;
	bidirectional_BFS();
	return 0;
}

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OPEN AT 2017.12.10

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