hdu 4021 24 Puzzle ( 逆序数判断是否可解 )
24 Puzzle
The ‘#’ denotes the positions that the tiles may be placed on. There are 24 possible positions in total, so one of them is not occupied by the tile. We can denote the empty position by zero.
Daniel could move the tiles to the empty position if the tile is on the top, bottom, left or right of the empty position. In this way Daniel can reorder the tiles on the board.
Usually he plays with this game by setting up a target states initially, and then trying to do a series of moves to achieve the target. Soon he finds that not all target states could be achieved.
He asks for your help, to determine whether he has set up an impossible target or not.
For each test case, the first line contains 24 integers denoting the initial states of the game board. The numbers are the describing the tiles from top to bottom, left to right. And the empty position is indicated by zero. You can assume that the number of each tile are different, and there must be exactly one empty position. The second line of test case also contains 24 integers denoting the target states.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
3 1 2 0 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
3 0 2 1 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
Y
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 30
#define MAXN 20005
#define mod 1000000007
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 0.000001
typedef long long ll;
using namespace std; int n,m,ans,p1,p2;
int a[maxn],b[maxn];
int mp[maxn]; bool presolve()
{
int i,j;
if(a[0]==0) swap(a[0],a[3]);
if(a[1]==0) swap(a[1],a[6]);
if(a[2]==0) swap(a[2],a[3]);
if(a[7]==0) swap(a[7],a[6]);
if(a[16]==0) swap(a[16],a[17]);
if(a[21]==0) swap(a[21],a[20]);
if(a[22]==0) swap(a[22],a[17]);
if(a[23]==0) swap(a[23],a[20]);
if(b[0]==0) swap(b[0],b[3]);
if(b[1]==0) swap(b[1],b[6]);
if(b[2]==0) swap(b[2],b[3]);
if(b[7]==0) swap(b[7],b[6]);
if(b[16]==0) swap(b[16],b[17]);
if(b[21]==0) swap(b[21],b[20]);
if(b[22]==0) swap(b[22],b[17]);
if(b[23]==0) swap(b[23],b[20]);
for(i=0;i<24;i++)
{
if(a[i]==0) p1=i;
if(b[i]==0) p2=i;
}
if(a[0]!=b[0]||a[1]!=b[1]||a[2]!=b[2]||a[7]!=b[7]||
a[16]!=b[16]||a[21]!=b[21]||a[22]!=b[22]||a[23]!=b[23])
return false ;
return true ;
}
int getstate(int x[])
{
int i,j,t,s,sum=0;
for(i=3;i<24;i++)
{
t=x[i];
if(t==0||i==7||i==16||i==21||i==22||i==23) continue ;
s=0;
for(j=i+1;j<24;j++)
{
if(x[j]==0||j==7||j==16||j==21||j==22||j==23) continue ;
if(x[j]<t) s++;
}
sum+=s;
}
return sum;
}
int main()
{
int i,j,t,s1,s2;
mp[3]=mp[4]=mp[5]=mp[6]=1;
mp[8]=mp[9]=mp[10]=mp[11]=2;
mp[12]=mp[13]=mp[14]=mp[15]=3;
mp[17]=mp[18]=mp[19]=mp[20]=4;
scanf("%d",&t);
while(t--)
{
for(i=0;i<24;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<24;i++)
{
scanf("%d",&b[i]);
}
if(presolve())
{
s1=getstate(a);
s2=getstate(b);
if((s1+s2)&1)
{
if(abs(mp[p1]-mp[p2])&1) printf("N\n");
else printf("Y\n");
}
else
{
if(!(abs(mp[p1]-mp[p2])&1)) printf("N\n");
else printf("Y\n");
}
}
else printf("Y\n");
}
return 0;
}
hdu 4021 24 Puzzle ( 逆序数判断是否可解 )的更多相关文章
- HDU 4911 Inversion (逆序数 归并排序)
Inversion 题目链接: http://acm.hust.edu.cn/vjudge/contest/121349#problem/A Description bobo has a sequen ...
- hdu 6048 Puzzle 拼图 逆序数
关于拼图和逆序数的关系可以看看这个 http://www.guokr.com/question/579400/ 然后求逆序数在判断就行了 按题意生成原始排列,观察发现,每一轮数后方比该数小的数的数量( ...
- HDU 1394 Minimum Inversion Number(最小逆序数/暴力 线段树 树状数组 归并排序)
题目链接: 传送门 Minimum Inversion Number Time Limit: 1000MS Memory Limit: 32768 K Description The inve ...
- HDU 4911 (树状数组+逆序数)
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4911 题目大意:最多可以交换K次,就最小逆序对数 解题思路: 逆序数定理,当逆序对数大于0时,若ak ...
- HDU 1394 Minimum Inversion Number(线段树/树状数组求逆序数)
Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java ...
- [HDU POJ] 逆序数
HDU 1394 Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/3276 ...
- 【归并排序】【逆序数】HDU 5775 Bubble Sort
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5775 题目大意: 冒泡排序的规则如下,一开始给定1~n的一个排列,求每个数字在排序过程中出现的最远端 ...
- HDU 1394 Minimum Inversion Number (线段树 单点更新 求逆序数)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1394 题意:给你一个n个数的序列,当中组成的数仅仅有0-n,我们能够进行这么一种操作:把第一个数移到最 ...
- hdu 1394(线段树) 最小逆序数
http://acm.hdu.edu.cn/showproblem.php?pid=1394 给出一列数组,数组里的数都是从0到n-1的,在依次把第一个数放到最后一位的过程中求最小的逆序数 线段树的应 ...
随机推荐
- 【数据结构】 最小生成树(四)——利用kruskal算法搞定例题×3+变形+一道大水题
在这一专辑(最小生成树)中的上一期讲到了prim算法,但是prim算法比较难懂,为了避免看不懂,就先用kruskal算法写题吧,下面将会将三道例题,加一道变形,以及一道大水题,水到不用高级数据结构,建 ...
- B/S架构和C/S架构
一.B/S架构 B/S结构(Browser/Server,浏览器/服务器模式),是WEB兴起后的一种网络结构模式,WEB浏览器是客户端最主要的应用软件.这种模式统一了客户端,将系统功能实现的核心部分集 ...
- ALL运算符
ALL在英文中的意思是“所有”,ALL运算符要求比较的值需要匹配子查询中的所有值.ALL运算符同样不能单独使用,必须和比较运算符共同使用. 下面的SQL语句用来检索在所有会员入会之前出版的图书: SE ...
- Hive 空指针(NPE)异常
空指针NullPointerException 1 Hive之前的一些BUG [HIVE-9430] - NullPointerException on ALTER TABLE ADD PARTITI ...
- [BZOJ4818][SDOI2017]序列计数(动规+快速幂)
4818: [Sdoi2017]序列计数 Time Limit: 30 Sec Memory Limit: 128 MBSubmit: 972 Solved: 581[Submit][Status ...
- BZOJ 2342 [Shoi2011]双倍回文(manacher+并查集)
[题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=2342 [题目大意] 记Wr为W串的倒置,求最长的形如WWrWWr的串的长度. [题解] ...
- insert失败自动执行update(duplicate先insert)
例如:有一张表 字段有 id主键自增,或者唯一索引:datetime时间 name名字 INSERT INTO TABLE (id,datetime) VALUES (1,1440000000), ...
- PHP 5.3.13 memcache win 64 配置和安装
--环境: windows 2008 R2 64位 wampserver2.2e-php5.3.13-httpd2.2.22-mysql5.5.24-x64 --目标: 实现 php 用memcach ...
- Eclipse错误导致无法启动The workspace exited with unsaved changes in the previous session
MyOpenSUSE:/home/jin/workspace # tail -f .metadata/.log !SESSION 2014-05-04 11:35:58.869 ----------- ...
- HDU 1507 Uncle Tom's Inherited Land*(二分匹配,输出任意一组解)
Uncle Tom's Inherited Land* Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (J ...