# infinity

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http://acm.pku.edu.cn/JudgeOnline/problem?id=1077

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`

Source

South Central USA 1998

`Source Code`
 Problem: 1077 User: lovecanon Memory: 9572K Time: 125MS Language: GCC Result: Accepted
• Source Code
• ```
#include<stdio.h>#include<string.h>#include<stdlib.h>struct node{    int state;    int pre;    int dir;}queue;int hash;int step;int a;int fac(int i){    switch(i)    {        case 0: return 1;        case 1: return 1;        case 2: return 2;        case 3: return 6;        case 4: return 24;        case 5: return 120;        case 6: return 720;        case 7: return 5040;        case 8: return 40320;    }    return 0;}int HASH(){    int i,j,k=0,b,ret=0,num=0;    for(i=0;i<3;i++)        for(j=0;j<3;j++)            b[k++]=a[i][j];    for(i=0;i<9;i++)    {        num=0;        for(j=0;j<i;j++)            if(b[j]>b[i])  num++;        ret+=fac(i)*num;    }    return ret;}void output(int len){    int i;    for(i=len;i>=0;i--)     {        if(step[i]==1) printf("l");        if(step[i]==2) printf("r");        if(step[i]==3) printf("u");        if(step[i]==4) printf("d");    }    printf("\n");}int main(){    char s;    int i,j,rear=0,front=0,tag=0;    rear++;    for(i=0;i<3;i++)        for(j=0;j<3;j++)        {            scanf("%s",s);            if(s=='x')  s='9';            queue[rear].state[i][j]=s-'0';        }    queue[rear].pre=0;queue[rear].dir=0;    for(i=0;i<3;i++)        for(j=0;j<3;j++)            a[i][j]=queue[rear].state[i][j];    hash[HASH()]=1;    while(front<rear)    {        int e,f,tmp,cntdir,len;        front++;        for(i=0;i<3;i++)            for(j=0;j<3;j++)            {                a[i][j]=queue[front].state[i][j];                if(a[i][j]==9) {e=i;f=j;}            }        if(f-1>=0)        {            cntdir=1;            tmp=a[e][f];            a[e][f]=a[e][f-1];            a[e][f-1]=tmp;            tmp=HASH();            if(tmp==0)             {                int t=front;                len=0;                step[len++]=cntdir;                while(queue[t].pre)                  {                    step[len++]=queue[t].dir;                    t=queue[t].pre;                }                output(len);                return 0;            }            if(!hash[tmp])             {                rear++;                for(i=0;i<3;i++)                    for(j=0;j<3;j++)                        queue[rear].state[i][j]=a[i][j];                queue[rear].dir=cntdir;queue[rear].pre=front;                hash[tmp]=1;            }            tmp=a[e][f];            a[e][f]=a[e][f-1];            a[e][f-1]=tmp;        }        if(f+1<3)        {            cntdir=2;            tmp=a[e][f];            a[e][f]=a[e][f+1];            a[e][f+1]=tmp;            tmp=HASH();            if(tmp==0)             {                int t=front;                len=0;                step[len++]=cntdir;                while(queue[t].pre)                  {                    step[len++]=queue[t].dir;                    t=queue[t].pre;                }                output(len);                return 0;            }            if(!hash[tmp])             {                rear++;                for(i=0;i<3;i++)                    for(j=0;j<3;j++)                        queue[rear].state[i][j]=a[i][j];                queue[rear].dir=cntdir;queue[rear].pre=front;                hash[tmp]=1;            }            tmp=a[e][f];            a[e][f]=a[e][f+1];            a[e][f+1]=tmp;        }        if(e-1>=0)        {            cntdir=3;            tmp=a[e][f];            a[e][f]=a[e-1][f];            a[e-1][f]=tmp;            tmp=HASH();            if(tmp==0)             {                int t=front;                len=0;                step[len++]=cntdir;                while(queue[t].pre)                  {                    step[len++]=queue[t].dir;                    t=queue[t].pre;                }                output(len);                return 0;            }            if(!hash[tmp])             {                rear++;                for(i=0;i<3;i++)                    for(j=0;j<3;j++)                        queue[rear].state[i][j]=a[i][j];                queue[rear].dir=cntdir;queue[rear].pre=front;                hash[tmp]=1;            }            tmp=a[e][f];            a[e][f]=a[e-1][f];            a[e-1][f]=tmp;        }        if(e+1<3)        {            cntdir=4;            tmp=a[e+1][f];            a[e+1][f]=a[e][f];            a[e][f]=tmp;            tmp=HASH();            if(tmp==0)             {                int t=front;                len=0;                step[len++]=cntdir;                while(queue[t].pre)                  {                    step[len++]=queue[t].dir;                    t=queue[t].pre;                }                output(len);                return 0;            }            if(!hash[tmp])             {                rear++;                for(i=0;i<3;i++)                    for(j=0;j<3;j++)                        queue[rear].state[i][j]=a[i][j];                queue[rear].dir=cntdir;queue[rear].pre=front;                hash[tmp]=1;            }            tmp=a[e+1][f];            a[e+1][f]=a[e][f];            a[e][f]=tmp;        }    }    printf("unsolvable\n");    return 0;}
```
posted on 2008-09-20 04:18 infinity 阅读(1959) 评论(0)  编辑 收藏 引用 所属分类: acm 只有注册用户登录后才能发表评论。 【推荐】超50万行VC++源码: 大型组态工控、电力仿真CAD与GIS源码库 相关文章: