Just an Old Puzzle (law) HDU - 6620

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.

InputThe 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.OutputFor 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

 

Meaning of the questions: should be well understood

Analysis: odd situation up to two digital game, when and only when the grid under two situations in the number sequence n * n-1 elements line once written (without regard to space), the number of reverse number the same odd number. Conclusions can also be extended to

n is an even number, and the two up situation, if and only if the write sequence, when the difference of the number of reverse and two spaces situation where the difference between the number of rows of the same parity. In summary, n * m digital solvable problem determination, can be converted to merge sort of the reverse seek to solve.

 

AC Code:

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn = 1e5+10;
 7 #define LL long long
 8 #define INF 0x3f3f3f3f
 9 int T,n;
10 int a[20];
11 
12 int main(){
13     scanf("%d",&T);
14     while(T--){
15         int res,ans=0;
16         for(int i=1;i<=16;i++){
17             scanf("%d",&a[i]);
18             if(!a[i]) res=i/4+(i%4?1:0);
19             else{
20                 for(int j=1;j<i;j++){
21                     if(!a[j]) continue;
22                     else if(a[j]>a[i]) ans++;
23                 }
24             }
25         }
26         if((4-res)%2==ans%2) printf("Yes\n");
27         else printf("No\n");
28     }
29 
30 
31     return 0;
32 }