搜索_IDA*_HDOJ3567_Eight II

Problem Description

Eight-puzzle, which is also called "Nine grids", comes from an old game.

In this game, you are given a 3 by 3 board and 8 tiles. The tiles are numbered from 1 to 8 and each covers a grid. As you see, there is a blank grid which can be represented as an 'X'. Tiles in grids having a common edge with the blank grid can be moved into that blank grid. This operation leads to an exchange of 'X' with one tile.

We use the symbol 'r' to represent exchanging 'X' with the tile on its right side, and 'l' for the left side, 'u' for the one above it, 'd' for the one below it.

A state of the board can be represented by a string S using the rule showed below.

The problem is to operate an operation list of 'r', 'u', 'l', 'd' to turn the state of the board from state A to state B. You are required to find the result which meets the following constrains:
1. It is of minimum length among all possible solutions.
2. It is the lexicographically smallest one of all solutions of minimum length.

Input

The first line is T (T <= 200), which means the number of test cases of this problem.

The input of each test case consists of two lines with state A occupying the first line and state B on the second line.
It is guaranteed that there is an available solution from state A to B.

Output

For each test case two lines are expected.

The first line is in the format of "Case x: d", in which x is the case number counted from one, d is the minimum length of operation list you need to turn A to B.
S is the operation list meeting the constraints and it should be showed on the second line.

思路分析:

本题中明确说明每个测试样例均有解, 且需计算字典序最小的移动方案, 故适合使用IDA*, 至于估价函数的设计参考本人另外两篇有关八数码问题的博客即可.

//HDOJ3567_Eight II
#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
const int dx[4] = {1, 0, 0, -1}, dy[4] = {0, -1, 1, 0};
const char dir[4] = {'d', 'l', 'r', 'u'};
int ndir[200];//ndir[i]:位置i的反方向 
//空格使用数字9表示, tpos[i][0]:tar中数字i所在行, tpos[i][1]:所在列 
int tar[4][4], tpos[10][2];//空格使用数字9表示, tpos[i][0] 
int beg[4][4];
int maxDeep;
//返回当前beg每个元素与tar中对应相等元素所在位置的曼哈顿距离之和 
int F(){
	int res = 0;
	for(int i = 1; i <= 3; ++i)
		for(int j = 1; j <= 3; ++j)		
			if(beg[i][j] != 9){
				int t = beg[i][j];
				res += abs(i - tpos[t][0]) + abs(j - tpos[t][1]);
			}
	return res;
}
bool dfs(vector<char> &path){
	int f = F(); if(f + path.size() > maxDeep) return false;
	if(path.size() == maxDeep) return true;
	int x, y;//空格所在位置
	for(int i = 1; i <= 3; ++i)
		for(int j = 1; j <= 3; ++j)
			if(beg[i][j] == 9){
				x = i, y = j; break;
			} 
	for(int i = 0; i <= 3; ++i){
		if(path.size() && path.back() == ndir[dir[i]]) continue;
		int a = x + dx[i], b = y + dy[i];
		if(a >= 1 && a <= 3 && b >= 1 && b <= 3){
			swap(beg[x][y], beg[a][b]), path.push_back(dir[i]);
			if(dfs(path)) return true;
			swap(beg[x][y], beg[a][b]), path.pop_back();	
		}
	}
	return false;
}
int main(){
	ndir['u'] = 'd', ndir['d'] = 'u', ndir['l'] = 'r', ndir['r'] = 'l';
	int T, sn = 0; scanf("%d", &T);
	while(++sn, T--){
		char s[100]; scanf("%s", s + 1);
		for(int i = 1, k = 1; i <= 3; ++i)
			for(int j = 1; j <= 3; ++j, ++k) 
				beg[i][j] = s[k] == 'X'? 9: s[k] - '0';
		scanf("%s", s + 1);
		for(int i = 1, k = 1; i <= 3; ++i)
			for(int j = 1; j <= 3; ++j, ++k){
					tar[i][j] = s[k] == 'X'? 9: s[k] - '0';
					int t = tar[i][j];
					tpos[t][0] = i, tpos[t][1] = j;
				}
		if(!F()){
			cout << "Case " << sn << ": " << 0 << endl;
			cout << endl; continue;
		}
		vector<char> path;
		maxDeep = 0; while(++maxDeep, !dfs(path));
		cout << "Case " << sn << ": " << path.size() << endl;
		for(int i = 0; i < path.size(); ++i) cout << path[i];
		cout << endl;
	}
	return 0;
}