HDOJ 题目1043 Eight（单向BFS，康拓展开，打表）

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16879    Accepted Submission(s): 4640
Special Judge

Problem 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 1213 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 1213 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, 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

 

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. Do not print a blank line between cases.

 

Sample Input

2  3  4  1  5  x  7  6  8

 

Sample Output

ullddrurdllurdruldr

 

Source

South Central USA 1998 (Sepcial Judge Module By JGShining)

 

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题目大意：

问序列变成12345678x的路径

ac代码

153941362015-11-05 17:54:45Accepted1043202MS8936K2668 BC++XY_

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>using namespace std;char num[15];int fac[15];void Fac(){    int i;    fac[1]=1;    for(i=2;i<=10;i++)        fac[i]=fac[i-1]*i;}int cot(char *num){    int i,j;    int sum=0;    for(i=0;i<8;i++)    {        int c=0;        for(j=i+1;j<9;j++)        {            if(num[i]>num[j])                c++;        }        sum+=c*fac[9-i-1];    }    return sum+1;}struct s{    int x,y;    int cnt;    char num[15];}a,temp;int vis[1010000];int dirx[4]={0,1,0,-1};int diry[4]={1,0,-1,0};char p[1010000];int pre[1010000];void bfs(){    int i;    queue<struct s>q;    memset(vis,0,sizeof(vis));    for(i=0;i<8;i++)        a.num[i]=i+'1';    a.num[8]='0';    a.cnt=cot(a.num);    //printf("%d\n",a.cnt);    a.x=2;    a.y=2;    q.push(a);    vis[a.cnt]=1;    p[a.cnt]=-1;    while(!q.empty())    {        a=q.front();        q.pop();        for(int i=0;i<4;i++)        {            temp.x=a.x+dirx[i];            temp.y=a.y+diry[i];            if(temp.x<0||temp.x>2||temp.y<0||temp.y>2)                continue;            int pos2=temp.x*3+temp.y;            int pos1=a.x*3+a.y;            strcpy(temp.num,a.num);            swap(temp.num[pos1],temp.num[pos2]);            temp.cnt=cot(temp.num);          //  printf("%d\n",temp.cnt);            if(vis[temp.cnt])                continue;            vis[temp.cnt]=1;            pre[temp.cnt]=a.cnt;            if(i==0)                p[temp.cnt]='l';            else                if(i==1)                    p[temp.cnt]='u';                else                    if(i==2)                        p[temp.cnt]='r';                    else                        p[temp.cnt]='d';            q.push(temp);        }    }}char str[1010];void print(int tt){    int x=tt;    while(p[x]!=-1)    {        printf("%c",p[x]);        x=pre[x];    }    printf("\n");}int main(){    Fac();    bfs();    while(gets(str)!=NULL)    {        int i;        int len=strlen(str);        char ss[15],k=0;        for(i=0;i<len;i++)        {            if(str[i]=='x')            {                ss[k++]='0';            }            else                if(str[i]>='1'&&str[i]<='8')                    ss[k++]=str[i];        }        ss[k]='\0';        int tt=cot(ss);        if(!vis[tt])        {            printf("unsolvable\n");        }        else        {            print(tt);        }    }}