首页 > ACM题库 > HDU-杭电 > HDU 1043 Eight-A*算法-[解题报告] C++
2013
11-26

HDU 1043 Eight-A*算法-[解题报告] C++

Eight

问题描述 :

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.

输入:

You will receive, several descriptions of configuration of the 8 puzzle. One 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

输出:

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. Do not print a blank line between cases.

样例输入:

2  3  4  1  5  x  7  6  8

样例输出:

ullddrurdllurdruldr

转载请注明出处,谢谢 http://blog.csdn.net/ACM_cxlove?viewmode=contents           by—cxlove

第一个A*搜索,A*是一种启发式搜索,g为已花代价,h为估计的剩余代价,而A*是根据f=g+h作为估价函数进行排列,也就是优先选择可能最优的节点进行扩展。

对于八数码问题,以下几个问题需要知道

判断有无解问题:根据逆序数直接判断有无解,对于一个八数码,依次排列之后,每次是将空位和相邻位进行调换,研究后会发现,每次调换,逆序数增幅都为偶数,也就是不改变奇偶性,所以只需要根据初始和目标状态的逆序数正负判断即可。

HASH问题:根据的是康托展开,具体证明请找网上资料

以及估价函数H:是根据与目标解的曼哈顿距离,也就是每个数字与目标位置的曼哈顿距离之和。

以了以上的基础,便可以通过A*解决八数码问题。

对于这题,实验了下,优先队列第一关键字为f,第二关键字为h,耗时2s+,第一关键字为f,第二关键字为g,耗时1s+,第一关键字为h,第二关键字为g,耗时450ms左右。在搜索过程中,加上判断是否有解,时间变化不大。POJ上0ms

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<string>
#define inf 1<<30
#define eps 1e-7
#define LD long double
#define LL long long
#define maxn 1000000005
using namespace std;
struct Node{
	int maze[3][3];   //八数码具体情况 
	int h,g;    //两个估价函数
	int x,y;   //空位的位置
	int Hash;   //HASH值
	bool operator<(const Node n1)const{     //优先队列第一关键字为h,第二关键字为g
		return h!=n1.h?h>n1.h:g>n1.g;
	}
	bool check(){    //判断是否合法
		if(x>=0&&x<3&&y>=0&&y<3)
			return true;
		return false;
	}
}s,u,v,tt;
int HASH[9]={1,1,2,6,24,120,720,5040,40320};   //HASH的权值
int destination=322560;   //目标情况的HASH值
int vis[400000];            //判断状态已遍历,初始为-1,否则为到达这步的转向
int pre[400000];        //路径保存
int way[4][2]={{0,1},{0,-1},{1,0},{-1,0}};   //四个方向
void debug(Node tmp){
	for(int i=0;i<3;i++){
		for(int j=0;j<3;j++)
			printf("%d ",tmp.maze[i][j]);
		printf("\n");
	}
	printf("%d %d\n%d %d\n",tmp.x,tmp.y,tmp.g,tmp.h);
	printf("hash=%d\n",tmp.Hash);
}
int get_hash(Node tmp){    //获得HASH值
	int a[9],k=0;
	for(int i=0;i<3;i++)
		for(int j=0;j<3;j++)
			a[k++]=tmp.maze[i][j];
	int res=0;
	for(int i=0;i<9;i++){
		int k=0;
		for(int j=0;j<i;j++)
			if(a[j]>a[i])
				k++;
		res+=HASH[i]*k;
	}
	return res;
}
bool isok(Node tmp){    //求出逆序对,判断是否有解
	int a[9],k=0;
	for(int i=0;i<3;i++)
		for(int j=0;j<3;j++)
			a[k++]=tmp.maze[i][j];
	int sum=0;
	for(int i=0;i<9;i++)
		for(int j=i+1;j<9;j++)
			if(a[j]&&a[i]&&a[i]>a[j])
				sum++;
	return !(sum&1);    //由于目标解为偶数,所以状态的逆序数为偶数才可行
}
int get_h(Node tmp){   //获得估价函数H
	int ans=0;
	for(int i=0;i<3;i++)
		for(int j=0;j<3;j++)
			if(tmp.maze[i][j])
				ans+=abs(i-(tmp.maze[i][j]-1)/3)+abs(j-(tmp.maze[i][j]-1)%3);
	return ans;
}
void astar(){    //搜索
	priority_queue<Node>que;
	que.push(s);
	while(!que.empty()){
		u=que.top();
		que.pop();
		for(int i=0;i<4;i++){
			v=u;
			v.x+=way[i][0];
			v.y+=way[i][1];
			if(v.check()){
				swap(v.maze[v.x][v.y],v.maze[u.x][u.y]);   //将空位和相邻位交换
				v.Hash=get_hash(v);			    //得到HASH值
				if(vis[v.Hash]==-1&&isok(v)){   //判断是否已遍历且是否可行,后者可以不要
					vis[v.Hash]=i;           //保存方向
					v.g++;;                  //已花代价+1
					pre[v.Hash]=u.Hash;     //保存路径
					v.h=get_h(v);           //得到新的估价函数H
					que.push(v);     //入队
				}
				if(v.Hash==destination)
					return ;
			}
		}
	}
}
void print(){
	string ans;
	ans.clear();
	int nxt=destination;
	while(pre[nxt]!=-1){  //从终点往起点找路径
		switch(vis[nxt]){   //四个方向对应
		case 0:ans+='r';break;
		case 1:ans+='l';break;
    	case 2:ans+='d';break;
    	case 3:ans+='u';break;
    	}
		nxt=pre[nxt];	
	}
	for(int i=ans.size()-1;i>=0;i--)
		putchar(ans[i]);
	puts("");
}
int main(){
	char str[100];
	while(gets(str)!=NULL){
		int k=0;
		memset(vis,-1,sizeof(vis));
		memset(pre,-1,sizeof(pre));
		for(int i=0;i<3;i++)
			for(int j=0;j<3;j++){
				if((str[k]<='9'&&str[k]>='0')||str[k]=='x'){
					if(str[k]=='x'){
						s.maze[i][j]=0;
						s.x=i;
						s.y=j;
					}
					else
						s.maze[i][j]=str[k]-'0';
				}
				else
					j--;
				k++;
			}
		if(!isok(s)){   //起始状态不可行
			printf("unsolvable\n");
			continue;
		}
		s.Hash=get_hash(s);
		if(s.Hash==destination){   //起始状态为目标状态
			puts("");
			continue;
		}
		vis[s.Hash]=-2;
		s.g=0;s.h=get_h(s);
		astar();
		print();
	}
	return 0;
}

 


  1. 嗯 分析得很到位,确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样:push时,比较要push的elem和辅助栈的栈顶,elem<=min.top(),则min.push(elem).否则只要push(elem)就好。在pop的时候,比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();},否则{stack.pop();}.

  2. “再把所有不和该节点相邻的节点着相同的颜色”,程序中没有进行不和该节点相邻的其他节点是否相邻进行判断。再说求出来的也不一样是颜色数最少的