标签：cas struct distrib following queue 通过 new ++i 需要
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem Description - 题目描述
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。
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
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++
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #define MX 362880 5 #define bitMX 10 6 7 int tre[bitMX]; 8 int lowBit(int a) { return -a&a; } 9 void add(int i) { 10 while (i < bitMX) { ++tre[i]; i += lowBit(i); } 11 } 12 int sum(int i) { 13 int opt = 0; 14 while (i) { opt += tre[i]; i -= lowBit(i); } 15 return opt; 16 } 17 18 int ktf[9], data[9]; 19 int preKte() { 20 int i, j, opt = 0; 21 memset(tre, 0, sizeof tre); 22 for (i = 8; ~i; --i) { 23 opt += sum(data[i])*ktf[i]; 24 add(data[i]); 25 } 26 return opt; 27 } 28 void kte(int a) { 29 int i, j, tmp[9]; 30 for (i = 0; i < 9; ++i) tmp[i] = i + 1; 31 for (i = 0; i < 8; ++i) { 32 j = a / ktf[i]; a %= ktf[i]; 33 data[i] = tmp[j]; 34 memcpy(tmp + j, tmp + j + 1, sizeof(int)*(8 - j)); 35 } 36 data[i] = tmp[0]; 37 } 38 39 int lst[MX]; 40 char pre[MX]; 41 void push(int now, int i, int j, char c, std::queue<int> &q) { 42 data[i] ^= data[j]; data[j] ^= data[i]; data[i] ^= data[j]; 43 int nxt = preKte(); 44 if (!pre[nxt]) { 45 lst[nxt] = now; pre[nxt] = c; 46 q.push(nxt); 47 } 48 data[i] ^= data[j]; data[j] ^= data[i]; data[i] ^= data[j]; 49 } 50 void init() { 51 int i, j, now, nxt; 52 ktf[7] = 1; 53 for (i = 6, j = 2; ~i; --i, ++j) ktf[i] = ktf[i + 1] * j; 54 memset(lst, -1, sizeof lst); 55 lst[0] = 0; pre[0] = ‘ ‘; 56 std::queue<int> q; q.push(0); 57 while (!q.empty()) { 58 now = q.front(); q.pop(); 59 kte(now); 60 for (i = 0; data[i] != 9; ++i); 61 if (i > 2) push(now, i, i - 3, ‘d‘, q); 62 if (i < 6) push(now, i, i + 3, ‘u‘, q); 63 if (i % 3) push(now, i, i - 1, ‘r‘, q); 64 if ((i + 1) % 3) push(now, i, i + 1, ‘l‘, q); 65 } 66 67 68 } 69 int main() { 70 init(); 71 int i, j; 72 char red[18]; 73 while (gets(red)) { 74 for (i = j = 0; i < 18; i += 2, ++j) data[j] = red[i] == ‘x‘ ? 9 : red[i] - ‘0‘; 75 if (~lst[i = preKte()]) { 76 for (j = i; j; j = lst[j]) putchar(pre[j]); 77 puts(""); 78 } 79 else puts("unsolvable"); 80 } 81 return 0; 82 }
标签：cas struct distrib following queue 通过 new ++i 需要
原文地址：http://www.cnblogs.com/Simon-X/p/6701038.html
