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自然万物都趋向从有序变得无序

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Eight HDU - 1043

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题目链接:点我

题目

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, 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

题意:

经典的八数码问题,问你要怎样变换才能从初始状态变到目标状态.

思路:

bfs + 康拓展开,紫书的一道例题,只是紫书没有记录路径.稍微变换一个即可

代码:

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#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};
struct p
{
    int a[9];
};
int last[362880];
int fx[362880];
int fact[9];
int vis[362880];
void init()
{
    fact[0]=1;
    for(int i=1;i<9;i++) fact[i]=fact[i-1]*i;
}
int codee(p x)
{
    int code=0;
    for(int i=0;i<9;i++)
    {
        int cnt=0;
        for(int j=i+1;j<9;j++)
            if (x.a[j]<x.a[i]) cnt++;
        code+=fact[8-i]*cnt;
    }
    return code;
}
void out(int x)
{
    if (last[x]==-1) return;
    switch (fx[x])
    {
        case 0:printf("d");break;
        case 1:printf("u");break;
        case 2:printf("r");break;
        case 3:printf("l");break;
    }
    out(last[x]);
}
int bfs()
{
    p b;
    for(int i=0;i<8;i++)
        b.a[i]=i+1;
    b.a[8]=0;
    int ha=codee(b);
    last[ha]=-1;
    vis[ha]=1;
    queue<p> que;
    que.push(b);
    while(!que.empty())
    {
        p u=que.front();que.pop();
        int hau=codee(u);
        int z;
        for(z=0;z<9;z++) if (u.a[z]==0) break;
        int x=z/3,y=z%3;
        for(int i=0;i<4;i++)
        {
            p v=u;
            int tx=x+dx[i],ty=y+dy[i],tz=tx*3+ty;
            if (tx>=0&&tx<3&&ty>=0&&ty<3)
            {
                swap(v.a[z],v.a[tz]);
                int haa=codee(v);
                if (!vis[haa])
                {
                    fx[haa]=i;
                    last[haa]=hau;
                 que.push(v);vis[haa]=1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    char c;
    init();
    memset(vis,0,sizeof(vis));
    bfs();
    while(scanf(" %c",&c)!=EOF)
    {
        p x;
        if (c!='x') x.a[0]=c-48;
        else x.a[0]=0;
        for(int i=1;i<9;i++)
        {
            scanf(" %c",&c);
            if (c!='x')x.a[i]=c-48;
            else x.a[i]=0;
        }
        int code=codee(x);
        if (vis[code])
        {
            out(code);
        }
        else printf("unsolvable");
        printf("\n");
    }
    return 0;
}