hdu 4021 24 Puzzle ( 逆序数判断是否可解 )

24 Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

Total Submission(s): 1306    Accepted Submission(s): 381

 

Problem Description

Daniel likes to play a special board game, called 24 puzzle. 24 puzzle is such a game that there are tiles with the number 1 to 23 in a play board like the follow picture:

  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.

 

 

Input

The first line of input contains an integer denoting the number of test cases.
  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.

 

 

Output

For each test case, if the target is impossible to achieve, output ‘Y’ in a single line, otherwise, output ‘N’.

 

 

Sample Input

2 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

 

 

Sample Output

N Y

 

 

Source

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

思路：

认真分析只有中间4*4的方格是有效的，其他8个方格可以去掉，因为只有0在旁边才可以移动，所以只需要将0移进4*4方格，判断里面的数是否相等不等肯定不行，然后就是逆序数问题了，上下交换逆序数改变，左右交换逆序数不变，两个状态的奇偶要与0所在的行的差的奇偶相同。

感想：

比赛时非常肯定这是逆序数问题，一直在纠结那多的八个方格，汗，连8个方格可以去掉都没分析出来，好菜呀！

代码：

#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;
}