UESTC 485 Game(康托展开,bfs打表)
Game
Time Limit: 4000/2000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
Submit Status
title
Today I want to introduce an interesting game to you. Like eight puzzle, it is a square board with 99 positions, but it filled by 99 numbered tiles. There is only one type of valid move, which is to rotate one row or column. That is, three tiles in a row or column are moved towards the head by one tile and the head tile is moved to the end of the row or column. So it has 1212 different moves just as the picture left. The objective in the game is to begin with an arbitrary configuration of tiles, and move them so as to get the numbered tiles arranged as the target configuration.
title
Now the question is to calculate the minimum steps required from the initial configuration to the final configuration. Note that the initial configuration is filled with a permutation of 11 to 99, but the final configuration is filled with numbers and * (which can be any number).
Input
The first line of input contains an integer TT (T≤1000T≤1000), which is the number of data sets that follow.
There are 66 lines in each data set. The first three lines give the initial configuration and the next three lines give the final configuration.
Output
For every test case, you should output Case #k: first, where kk indicates the case number and starts at 11. Then the fewest steps needed. If he can’t move to the end, just output No Solution! (without quotes).
Sample input and output
Sample Input Sample Output
2
1 2 3
4 5 6
7 8 9
1 2 3
4 5 6
7 9 8
1 2 3
4 5 6
7 8 9
8 * 9
5 3 7
2 * *
Case #1: No Solution!
Case #2: 7
利用康托展开进行bfs预处理。题目给的一个起始的九宫格,和一个目标的九宫格。 不能直接用目标的九宫格去找起始的九宫格,会超时,应该根据把起始九宫格当作
1 2 3
4 5 6
7 8 9
然后确定目标九宫格是怎么样的,这样就可以直接用之前打的表了。预处理就是处理1 2 3 4 5 6 7 8 9到每种九宫格的步数
关于康托展开,给出一篇博文吧
http://blog.csdn.net/dacc123/article/details/50952079
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <queue>
using namespace std;
struct Node
{
int a[5][5];
int sta;
};
queue<Node> q;
int b[10];
int fac[10];
int vis[400000];
int pre[400000];
int ans;
int f1[10];
int f2[10];
int tran[10];
char ch[10];
bool used[10];
Node cyk;
void facfun()
{
fac[0]=1;
for(int i=1;i<=9;i++)
{
fac[i]=i*fac[i-1];
}
}
int kt(Node q)
{
int cnt=0;
for(int i=1;i<=3;i++)
for(int j=1;j<=3;j++)
b[++cnt]=q.a[i][j];
int sum=0,num=0;
for(int i=1;i<=9;i++)
{
num=0;
for(int j=i+1;j<=9;j++)
{
if(b[i]>b[j])
num++;
}
sum+=num*fac[9-i];
}
return sum;
}
void bfs(Node t)
{
q.push(t);
vis[t.sta]=1;
pre[t.sta]=0;
while(!q.empty())
{
Node term=q.front();
q.pop();
for(int i=1;i<=12;i++)
{
Node temp=term;
if(i<=3)
{
temp.a[i][1]=term.a[i][3];
temp.a[i][2]=term.a[i][1];
temp.a[i][3]=term.a[i][2];
}
else if(i>3&&i<=6)
{
temp.a[i-3][1]=term.a[i-3][2];
temp.a[i-3][2]=term.a[i-3][3];
temp.a[i-3][3]=term.a[i-3][1];
}
else if(i>6&&i<=9)
{
temp.a[1][i-6]=term.a[3][i-6];
temp.a[2][i-6]=term.a[1][i-6];
temp.a[3][i-6]=term.a[2][i-6];
}
else if(i>9&&i<=12)
{
temp.a[1][i-9]=term.a[2][i-9];
temp.a[2][i-9]=term.a[3][i-9];
temp.a[3][i-9]=term.a[1][i-9];
}
int state=kt(temp);
if(vis[state])
continue;
temp.sta=state;
vis[state]=1;
pre[state]=pre[term.sta]+1;
q.push(temp);
}
}
}
void init()
{
memset(vis,0,sizeof(vis));
memset(pre,-1,sizeof(pre));
facfun();
Node st;int cnt=0;
for(int i=1;i<=3;i++)
for(int j=1;j<=3;j++)
st.a[i][j]=++cnt;
st.sta=0;
bfs(st);
}
int anspos;
void dfs(int i)
{
if(i==10)
{
/*for(int p=1;p<=3;p++)
{
for(int k=1;k<=3;k++)
{
cout<<cyk.a[p][k]<<" ";
}
cout<<endl;
}*/
int c=pre[kt(cyk)];
if(c==-1) return;
ans=min(ans,c);return;
}
if(f2[i]==0)
{
for(int j=1;j<=9;j++)
{
if(!used[j])
{
used[j]=true;
int y=i%3,x;
if(y==0){x=i/3;y=3;}
else {x=i/3+1;}
cyk.a[x][y]=j;
dfs(i+1);
used[j]=false;
}
}
}
else
{
int y=i%3,x;
if(y==0){x=i/3;y=3;}
else {x=i/3+1;}
cyk.a[x][y]=f2[i];
dfs(i+1);
}
}
int main()
{
int t;
scanf("%d",&t);
init();
int cas=0;
while(t--)
{
memset(used,0,sizeof(used));
for(int i=1;i<=9;i++)
{
scanf("%d",&f1[i]);
tran[f1[i]]=i;
}
for(int i=1;i<=9;i++)
{
scanf("%s",ch);
f2[i]=ch[0]-'0';
if(f2[i]>=1&&f2[i]<=9)
f2[i]=tran[f2[i]],used[f2[i]]=true;
else
f2[i]=0;
}
ans=1000000;
dfs(1);
if(ans>=1000000)
printf("Case #%d: No Solution!\n",++cas);
else
printf("Case #%d: %d\n",++cas,ans);
}
return 0;
}
UESTC 485 Game(康托展开,bfs打表)的更多相关文章
- [HDOJ1043]Eight(康托展开 BFS 打表)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1043 八数码问题,因为固定了位置所以以目标位置开始搜索,把所有情况(相当于一个排列)都记录下来,用康托 ...
- HDU 3567 Eight II 打表,康托展开,bfs,g++提交可过c++不可过 难度:3
http://acm.hdu.edu.cn/showproblem.php?pid=3567 相比Eight,似乎只是把目标状态由确定的改成不确定的,但是康托展开+曼哈顿为h值的A*和IDA*都不过, ...
- HDU 1043 & POJ 1077 Eight(康托展开+BFS+预处理)
Eight Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 30176 Accepted: 13119 Special ...
- HDU 1043 & POJ 1077 Eight(康托展开+BFS | IDA*)
Eight Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 30176 Accepted: 13119 Special ...
- HDU 1430 魔板(康托展开+BFS+预处理)
魔板 Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submis ...
- 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 ...
- poj1077(康托展开+bfs+记忆路径)
题意:就是说,给出一个三行三列的数组,其中元素为1--8和x,例如: 1 2 3 现在,需要你把它变成:1 2 3 要的最少步数的移动方案.可以右移r,左移l,上移u,下移dx 4 6 4 5 67 ...
- HDU1043 八数码(BFS + 打表)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1043 , 康托展开 + BFS + 打表. 经典八数码问题,传说此题不做人生不完整,关于八数码的八境界 ...
- 转换地图 (康托展开+预处理+BFS)
Problem Description 在小白成功的通过了第一轮面试后,他来到了第二轮面试.面试的题目有点难度了,为了考核你的思维能量,面试官给你一副(2x4)的初态地图,然后在给你一副(2x4)的终 ...
随机推荐
- imx6dl uboot 移植
新版的BSP引进的设备树的机制,在uboot中还添加了menuconfig的配置菜单. 参考官网的文档进行uboot移植,本文使用的cpu是imx6dl,uboot版本2015.04. 我要添加一个名 ...
- e591. Drawing Simple Text
See also e575 The Quintessential Drawing Program. public void paint(Graphics g) { // Set the desired ...
- Ubuntu 14.04 安装R 环境
Introduction R is a popular open source programming language that specializes in statistical computi ...
- Qt 定时器Timer使用
From: http://dragoon666.blog.163.com/blog/static/107009194201092602326598/ 1.新建Gui工程,在主界面上添加一个标签labe ...
- 《开源框架那些事儿22》:UI框架设计实战
UI是User Interface的缩写.通常被觉得是MVC中View的部分,作用是提供跟人机交互的可视化操作界面. MVC中Model提供内容给UI进行渲染,用户通过UI框架产生响应,一般而言会由控 ...
- ftp命令行工具如何 连接 非标准21端口(其他端口)的ftp服务器
windows: step1:ftp命令进入ftp交互环境 step2:ftp>open ip空格port 然后...
- 阮一峰---javascript系列
2013.05.11:如何做到 jQuery-free?(29条评论) 2013.01.23:JavaScript Source Map 详解(14条评论) 2013.01.14:Javascript ...
- GIS-007-Terrain跨域访问
方法一: 在数据服务目录中添加一个Web.config文件,文件内容是: <?xml version="1.0" encoding="UTF-8"?> ...
- python2.0_s12_day14_jQuery详解
jquery的中文介绍文档:http://www.php100.com/manual/jquery/jQuery之基本选择器jQuery中提供的用于获取标签的方法都有哪些? jQuery提供的 &qu ...
- 什么是LTE?
LTE是英文Long Term Evolution的缩写.LTE也被通俗的称为3.9G,具有100Mbps的数据下载能力,被视作从3G向4G演进的主流技术.它改进并增强了3G的空中接入技术,采用OFD ...