2019 Multi-University Training Contest 4 Just an Old Puzzle hdu6620 （结论题？）

Just an Old Puzzle

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 266    Accepted Submission(s): 176

 

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

 

 

Source

2019 Multi-University Training Contest 4

 

图片转自：https://blog.youkuaiyun.com/birdmanqin/article/details/97953000

AC代码：

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
using namespace std;
typedef long long ll;
const int MAXN = 50;
const int INF = 0x3f3f3f3f;
ll t, n;
int arr[MAXN];
int main() {
	//ios::sync_with_stdio(false);

    scanf("%lld", &t);
    int blank;
    int castle;
	while(t--){
        castle = 0;
        rep(i, 1, 16){
            scanf("%d", &arr[i]);
            if(arr[i] == 0)
                blank = (i-1)/4+1;
        }
        rep(i, 1, 16){
            if(arr[i] == 0) continue;
            rep(j, i+1, 16){
                if(arr[j] == 0) continue;
                if(arr[i] > arr[j]) castle++;
            }
        }
        printf("%s\n", ((blank%2 == castle%2)?"Yes":"No"));
	}


	return 0;
}