HDU 1043 Eight（八数码）

HDU 1043 Eight（八数码）

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

 

Problem Description - 题目描述

　　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:

15-puzzle有着超过100年的历史；就算没听过，估计也见过。它由15片滑块组成，各块标有数字1到15，并且全都装在一个4 x 4的边框里，防止哪块突然丢了。空块为’x’；这道题的目标是排列滑块使之顺序如下：

 

 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:

唯一的合法操作是将’x’与其共边的滑块交换。例子如下，通过一系列移动解决一个被稍微打乱的问题。

 

 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.

上一行的字母表示每次’x’与哪块相邻的滑块交换；有效值'r'，'l'，'u' 和 'd'分别表示右、左、上、下。

并非所有情况都有解；在1870年，一个叫Sam Loyd的人就以发布了一个无解版本而出名，成功地难住了许多人。实际上，要制造无解的情况只需交换两个滑块（当然是非’x’滑块）。

这个问题中，你需要写个程序解决著名的八数码问题，滑块为三行三列。

 

Input - 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

你会得到若干个八数码的配置描述。每个描述都是一个滑块初始位置的列表，从上往下，从左往右，使用数字1到8，还有’x’表示。比如，

1 2 3 
x 4 6 
7 5 8 

描述如下：

1 2 3 x 4 6 7 5 8

Output - 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.

如果无解，输出”unsolvable''”，否则输出一个仅由'r', 'l', 'u' 和 'd' 组成的字符串，描述求解的步骤。这个字符串不含空格且单独在一行。用例间别输出空行。

 

Sample Input - 输入样例

2  3  4  1  5  x  7  6  8

 

Sample Output - 输出样例

ullddrurdllurdruldr

 

题解
　　逼你学新知识系列……（某废渣作死爆内存了……）
　　状态可以用全排列表示，用康托展开来压缩状态和去重，然后剩下的只有BFS了……（无聊用了树状数组求逆序数）
　　多组输入有点坑……其实如果符合情况直接输出空行也是可以的。

 

代码 C++

#include <cstdio>
#include <cstring>
#include <queue>
#define MX 362880
#define bitMX 10

int tre[bitMX];
int lowBit(int a) { return -a&a; }
void add(int i) {
    while (i < bitMX) { ++tre[i]; i += lowBit(i); }
}
int sum(int i) {
    int opt = 0;
    while (i) { opt += tre[i]; i -= lowBit(i); }
    return opt;
}

int ktf[9], data[9];
int preKte() {
    int i, j, opt = 0;
    memset(tre, 0, sizeof tre);
    for (i = 8; ~i; --i) {
        opt += sum(data[i])*ktf[i];
        add(data[i]);
    }
    return opt;
}
void kte(int a) {
    int i, j, tmp[9];
    for (i = 0; i < 9; ++i) tmp[i] = i + 1;
    for (i = 0; i < 8; ++i) {
        j = a / ktf[i]; a %= ktf[i];
        data[i] = tmp[j];
        memcpy(tmp + j, tmp + j + 1, sizeof(int)*(8 - j));
    }
    data[i] = tmp[0];
}

int lst[MX];
char pre[MX];
void push(int now, int i, int j, char c, std::queue<int> &q) {
    data[i] ^= data[j]; data[j] ^= data[i]; data[i] ^= data[j];
    int nxt = preKte();
    if (!pre[nxt]) {
        lst[nxt] = now; pre[nxt] = c;
        q.push(nxt);
    }
    data[i] ^= data[j]; data[j] ^= data[i]; data[i] ^= data[j];
}
void init() {
    int i, j, now, nxt;
    ktf[7] = 1;
    for (i = 6, j = 2; ~i; --i, ++j) ktf[i] = ktf[i + 1] * j;
    memset(lst, -1, sizeof lst);
    lst[0] = 0; pre[0] = ' ';
    std::queue<int> q; q.push(0);
    while (!q.empty()) {
        now = q.front(); q.pop();
        kte(now);
        for (i = 0; data[i] != 9; ++i);
        if (i > 2) push(now, i, i - 3, 'd', q);
        if (i < 6) push(now, i, i + 3, 'u', q);
        if (i % 3) push(now, i, i - 1, 'r', q);
        if ((i + 1) % 3) push(now, i, i + 1, 'l', q);
    }

    
}
int main() {
    init();
    int i, j;
    char red[18];
    while (gets(red)) {
        for (i = j = 0; i < 18; i += 2, ++j) data[j] = red[i] == 'x' ? 9 : red[i] - '0';
        if (~lst[i = preKte()]) {
            for (j = i; j; j = lst[j]) putchar(pre[j]);
            puts("");
        }
        else puts("unsolvable");
    }
    return 0;
}