HDU 1043 Eight (BFS·八数码·康托展开)
题意 输出八数码问题从给定状态到12345678x的路径
用康托展开将排列相应为整数 即这个排列在全部排列中的字典序 然后就是基础的BFS了
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5, M = 9;
int x[4] = { -1, 1, 0, 0};
int y[4] = {0, 0, -1, 1};
int fac[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};
int puz[N][M], nex[N], dir[N], vis[N], q[N]; int getCantor(int a[]) //康托展开 将排列转化为整数
{
int ret = 0;
for(int i = 0; i < M; ++i)
{
for(int j = i + 1; j < M; ++j)
if(a[j] < a[i]) ret += fac[M - i - 1];
}
return ret;
} void bfs()
{
int t[M] = {1, 2, 3, 4, 5, 6, 7, 8, 0};
int id = getCantor(t);
dir[id] = -1; memcpy(puz[id], t, sizeof(t));
memset(vis, 0, sizeof(vis)); int r, c, k, nr, nc, nk, nid;
int front = 0, rear = 0;
q[rear++] = id;
vis[id] = 1; while(front < rear)
{
int id = q[front++];
memcpy(t, puz[id], sizeof(t));
for(k = 0; t[k]; ++k); //找0的位置
r = k / 3, c = k % 3; //一维转二维 for(int i = 0; i < 4; ++i)
{
nr = r + x[i], nc = c + y[i], nk = nr * 3 + nc; if(nr < 0 || nr > 2 || nc < 0 || nc > 2) continue;
swap(t[k], t[nk]);
nid = getCantor(t);
memcpy(puz[nid], t, sizeof(t));
swap(t[k], t[nk]); if(vis[nid]) continue;
vis[nid] = 1;
q[rear++] = nid;
nex[nid] = id;
dir[nid] = i;
}
}
} int main()
{
char t[5], sdir[] = "durl";
int s[M], id;
bfs(); while(~scanf("%s", t))
{
s[0] = t[0] == 'x' ? 0 : t[0] - '0';
for(int i = 1; i < M; ++i)
{
scanf("%s", t);
s[i] = t[0] == 'x' ? 0 : t[0] - '0';
} id = getCantor(s);
if(!vis[id]) puts("unsolvable");
else
{
while(dir[id] >= 0)
{
printf("%c", sdir[dir[id]]);
id = nex[id];
}
puts("");
}
}
return 0;
}
//Last modified : 2015-07-05 11:15
Eight
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.
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
and start at the beginning of the line. Do not print a blank line between cases.
2 3 4 1 5 x 7 6 8
ullddrurdllurdruldr
HDU 1043 Eight (BFS·八数码·康托展开)的更多相关文章
- HDU 1043 Eight 【经典八数码输出路径/BFS/A*/康托展开】
本题有写法好几个写法,但主要思路是BFS: No.1 采用双向宽搜,分别从起始态和结束态进行宽搜,暴力判重.如果只进行单向会超时. No.2 采用hash进行判重,宽搜采用单向就可以AC. No.3 ...
- hdu1043Eight (经典的八数码)(康托展开+BFS)
建议先学会用康托展开:http://blog.csdn.net/u010372095/article/details/9904497 Problem Description The 15-puzzle ...
- HDU 3567 Eight II(八数码 II)
HDU 3567 Eight II(八数码 II) /65536 K (Java/Others) Problem Description - 题目描述 Eight-puzzle, which is ...
- Eight HDU - 1043 (双向BFS)
记得上人工智能课的时候老师讲过一个A*算法,计算估价函数(f[n]=h[n]+g[n])什么的,感觉不是很好理解,百度上好多都是用逆向BFS写的,我理解的逆向BFS应该是从终点状态出发,然后把每一种状 ...
- HDU 1043 Eight BFS
题意:就是恢复成1,2,3,4,5,6,7,8,0; 分析:暴力BFS预处理,所有路径,用康拓展开判重,O(1)打印 93ms 还是很快的 #include <iostream> #inc ...
- BFS:八数码问题
#include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> ...
- POJ 1077 Eight (BFS+康托展开)详解
本题知识点和基本代码来自<算法竞赛 入门到进阶>(作者:罗勇军 郭卫斌) 如有问题欢迎巨巨们提出 题意:八数码问题是在一个3*3的棋盘上放置编号为1~8的方块,其中有一块为控制,与空格相邻 ...
- hdu 1043 Eight (八数码问题)【BFS】+【康拓展开】
<题目链接> 题目大意:给出一个3×3的矩阵(包含1-8数字和一个字母x),经过一些移动格子上的数后得到连续的1-8,最后一格是x,要求最小移动步数. 解题分析:本题用BFS来寻找路径,为 ...
- hdu 1043 Eight 经典八数码问题
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1043 The 15-puzzle has been around for over 100 years ...
随机推荐
- Android 开发笔记___shape
shape_oval <?xml version="1.0" encoding="utf-8"?> <shape xmlns:android= ...
- 全局作用域 eval
eval是在caller的作用域里运行传给它的代码: var x = 'outer'; (function() { var x = 'inner'; eval('x'); // & ...
- 使用SQLPLUS创建用户名和表空间
用sqlplus为oracle创建用户和表空间用sqlplus为oracle创建用户和表空间用Oracle10g自带的企业管理器或PL/SQL图形化的方法创建表空间和用户以及分配权限是相对比较简单的, ...
- 解决ajax的parsererror错误的终极办法(后台传给前台的数据json问题)
解决ajax的parsererror错误的终极办法(后台传给前台的数据json问题) 出现这个问题的原因是因为后台传给前台的数据出现了问题,ajax对于json的格式特别的严格 下面是会出现这个问题的 ...
- 2017/10/10 jar包错误
Description Resource Path Location Type Archive for required library: 'WebContent/WEB-IN ...
- scala(一)Nothing、Null、Unit、None 、null 、Nil理解
相对于java的类型系统,scala无疑要复杂的多!也正是这复杂多变的类型系统才让OOP和FP完美的融合在了一起! Nothing: 如果直接在scala-library中搜索Nothing的话是找不 ...
- TensorFlow简易学习[1]:基本概念和操作示例
简介 TensorFlow是一个实现机器学习算法的接口,也是执行机器学习算法的框架.使用数据流式图规划计算流程,可以将计算映射到不同的硬件和操作系统平台. 主要概念 TensorFlow的计算可以表示 ...
- 手写particles
var canvas = document.getElementById('canvas'); var ctx = canvas.getContext('2d'); var Grewer = { in ...
- 利用docker搭建spark hadoop workbench
目的 用docker实现所有服务 在spark-notebook中编写Scala代码,实时提交到spark集群中运行 在HDFS中存储数据文件,spark-notebook中直接读取 组件 Spark ...
- javafx 聊天室WeChat
[toc] 功能和特性 基于socket实现的c/s架构的的通信 服务器和客户心跳连接 gson实现的消息通信机制 注册及登录 支持私聊和群聊. 动态更新用户列表以及用户消息提示 支持emoji表情, ...