Just an Old Puzzle (2019 Multi - schools, 1007)

Problem Description

You are given a 4 × 4 grid, which consists of 15 number cells and an empty cell.
All numbers are unique and ranged from 1 to 15.
In this board, the cells which are adjacent with the empty cell can move to the empty cell.
Your task is to make the input grid to the target grid shown in the figure below.
In the following example (sample input), you can get the target grid in two moves.

 

 

Input

The first line contains an integer T (1 <= T <= 10^5) denoting the number of test cases.
Each test case consists of four lines each containing four space-separated integers, denoting the input grid. 0 indicates the empty cell.

 

 

Output

For each test case, you have to print the answer in one line.
If you can’t get the target grid within 120 moves, then print 'No', else print 'Yes'.

 

 

Sample Input

2 1 2 3 4 5 6 7 8 9 10 0 12 13 14 11 15 1 2 3 4 5 6 7 8 9 10 11 12 13 15 14 0

 

 

Sample Output

Yes No

 Conclusion: Jigsaw solvable problems: the parity of the current line spacing of blank or whether the current value of the parity of the number of reverse sequence to determine

Code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int a[25];
int cc[25];
int main()
{
    int T;
    cin>>T;
    int k=0;

    while(T--)
    {
        int cnt=0;
        for(int t=0;t<4;t++)
        {
            for(int j=0;j<4;j++)
            {
                scanf("%d",&a[t*4+j]);
                if(a[t*4+j]!=0)
                {
                    cc[cnt++]=a[t*4+j];
                }
                if(a[t*4+j]==0)
                {
                    k=t;
                }
            }
        }
    int sum=0;
    for(int t=0;t<15;t++)
    {
        for(int j=t+1;j<15;j++)
        {
            if(cc[j]<cc[t])
            {
                sum++;
            }
        }
    }
    
    if(((3-k)%2)^(sum%2)==0)
    {
        puts("Yes");
    }
    else
    {
        puts("No");
    }    
    }
    return 0;
}