2019HDU多校训练第四场G-Just an Old Puzzle

Just an Old Puzzle

时间限制: 1 Sec  内存限制: 128 MB

题目描述

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.

输入

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.

输出

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'.

样例输入

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

   样例输出

Yes
No

代码

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int arr[30];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=1;i<=4;i++) scanf("%d",&arr[i]);
        for(int i=8;i>=5;i--)scanf("%d",&arr[i]);
        for(int i=9;i<=12;i++)scanf("%d",&arr[i]);
        for(int i=16;i>=13;i--)scanf("%d",&arr[i]);
        int cnt=0;
        for(int i=16;i>=1;i--)
            for(int j=i-1;j>=1;j--)
                if(arr[i]<arr[j]&&arr[i]!=0)
                    cnt++;
        if(cnt&1)printf("Yes
");
        else printf("No
");
    }
    return 0;
}