本文主要是介绍POJ 1077 Eight，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
链接：http://poj.org/problem?id=1077
题目：
| Language: Eight
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 | ||||||||||
解题思路：
这是一个经典的8数码问题，对于这个问题，我们可以用搜索的方法来解决。用bfs（）来找一条最短的可行路。对于这个问题，我们有几点要注意一下：1.这个问题的状态是9个数字（我们把x看成0）的一个排列，每种排列都代表一种状态，用bfs（）搜索，就是从当前的一个排列，转移到另一个可行的排列，我们可以用一个结构体来表示这种状态；2.说到bfs（），我们自然就会想到vis[]数组，标志当前的一个状态是否有访问过，防止重复访问，算是剪枝吧，对于这个问题，一个状态有9个数字，最直接的想法是开一个9维的数组，这样显然是不太可行的，会爆内存，要解决这个问题，我们可以用hash的方法，将9个数字表示的一个状态映射成一个数字，这样一共就只有9！种不同的状态了，对于这个hash，有一种比较经典的做法，康拓展开，大意是这样的，hash = a * 8！+ b * 7! + c * 6! + d * 5! + e * 4! + f * 3！+ g * 2! + h * 1! + i * 0!;其中a表示的是，从第一个数字开始往后，比第一个数字小的数的个数，b表示，从第二个数字开始往后，比第二个数字小的数的个数，依次类推；3.要记录当前的路径，我们可以开一个全局数组path，一个全局数组pre，分别记录，到当前状态hash值的路径，以及前一个状态的hash值；4.还有一点，最好用G++编译器提交，用C++会超时，估计是因为G++对stl有优化。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;struct P
{int step;char c;int a[9];
};
P state, dest;
char dir[5] = "udlr";
bool vis[362888];
int pre[362888];
char path[362888];
char ans[362888];
int dx[4] = {-1, 1, 0, 0};
int dy[4] = { 0, 0, -1, 1};
int f[10], step, h;bool fun(P a, P b)
{for(int i = 0; i < 9; i++){if(a.a[i] != b.a[i]) return false;}return true;
}int hash(P p)
{int ret = 0;for(int i = 0; i < 9; i++){int num = 0;for(int j = i + 1; j < 9; j++){if(p.a[j] < p.a[i]) num++;}ret += f[8-i] * num;}return ret;
}int bfs()
{memset(vis, false, sizeof(vis));queue<P> q;q.push(state);while(!q.empty()){P src = q.front(); q.pop();if(vis[hash(src)]) continue;vis[hash(src)] = true;int x, y, z, t;for(int i = 0; i < 9; i++){if(0 == src.a[i]){t = i;break;}}x = t / 3; y = t % 3;for(int i = 0; i < 4; i++){int tx = x + dx[i], ty = y + dy[i];if(tx >= 0 && tx <= 2 && ty >= 0 && ty <= 2){P p;for(int j = 0; j < 9; j++){p.a[j] = src.a[j];}z = tx * 3 + ty;swap(p.a[t], p.a[z]);if(vis[hash(p)]) continue;p.step = src.step + 1;pre[hash(p)] = hash(src);path[hash(p)] = dir[i];if(fun(p, dest)){step = p.step;h = hash(p);return 1;}q.push(p);}}}return 0;
}int main()
{for(int i = 0; i < 8; i++) dest.a[i] = i + 1;dest.a[8] = 0;f[0] = 1;for(int i = 1; i < 10; i++) f[i] = i * f[i-1];char c; int i = 0;while(c = getchar()){if(c >= '1' && c <= '8'){state.a[i++] = c - '0';}if('x' == c){state.a[i++] = 0;}if(9 == i) break;}state.step = 0;if(bfs()){for(int i = 0; i < step; i++){ans[i] = path[h];h = pre[h];}for(int i = step - 1; i >= 0; i--){printf("%c", ans[i]);}puts("");}else{puts("unsolvable");}return 0;
}这篇关于POJ 1077 Eight的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
