POJ 1077 八数码 BFS +A*（复习的时候看过来！这道题值的看啊啊啊~！！！）

D - Eight

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

题目大意：自己看啦

提交情况：T T T TT T T T T T T T T T T TT T AC...

目前只是用了BFS做了，有空要把A*补上

该说的都在注释里面

这里说一句，如果循环很多次的话，一个微小的声明的区别都会导致250ms的差别，

只能说姿势很重要，如果姿势不过关只能考研自己的人品看看能不能卡着数据过了

下面是BFS的AC代码

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<vector>#include<cmath>#define aim 0using namespace std;bool cantor[400000];int fac[10]={1,1,2,6,24,120,720,5040,40320,362880};//0!1!2!3!4!  5!  6!   7!  8!     9!int xd[4]={0,0,-1,1},yd[4]={-1,1,0,0};   //  移动方向上下左右，与下面的dir相意义对应char dir[4]={'u','d','l','r'};    //上下左右struct node{int s[10];int pos,a,b;int state;string str;node(){        memset(s,0,sizeof(s));    }}start,e;int hashint(int C[]){    int i,j;    int sum=0;    for(i=0;i<9;i++){        int res=0;        for(j=i+1;j<9;j++){            if(C[j]<C[i]){                res++;            }        }        sum+=res*fac[8-i];    }    return sum;}int main(){    int x;    char ch;    for(int k=0;k<9;){        scanf("%c",&ch);        if(ch=='x'){            start.s[k]=9;            start.pos=k++;        }        else if(ch<='9'&&ch>='1'){            start.s[k++]=ch-'0';        }    }    start.a=start.pos%3;    start.b=start.pos/3;    memset(cantor,false,sizeof(cantor));    start.state=hashint(start.s);    cantor[start.state]=1;    queue<node> q;    q.push(start);    while(!q.empty()){        node cur=q.front();        /*这道题果真还是应该用A*来解，        单纯bfs的时间卡得非常的紧，        牛神拿了我原来T的代码看了之后就按照我的思路敲了一遍结果AC了，        经过我无数遍的比对，发现问题就在这一句，如果使用这一句来申明队列，        那么提交时间是750ms，如果使用下面那两句，则是T，        这个真的是看人品...只能说出题人是想要卡bfs的只是卡得还不够紧。        看来还是姿势第一啊。没有姿势就只能赌人品了        */        /*node cur;        cur=q.front();*/        q.pop();        for(int i=0;i<4;i+=1){            int x,y;            x=cur.a+xd[i],y=cur.b+yd[i];//数组不越界            if(x<3&&y<3&&x>=0&&y>=0){                 node tempt=cur;                 tempt.s[y*3+x]=cur.s[cur.b*3+cur.a];                 tempt.s[cur.b*3+cur.a]=cur.s[y*3+x];                 tempt.state=hashint(tempt.s);                 if(!cantor[tempt.state]){                     cantor[tempt.state]=true;                     tempt.str=cur.str+dir[i];                     tempt.a=x;                     tempt.b=y;                     if(tempt.state==aim){                        e=tempt;                        cout<<e.str<<endl;                        return 0;                     }                     q.push(tempt);                  }             }        }    }    cout<<"unsolvable"<<endl;    return 0;}