The Eight Puzzle, among other sliding-tile puzzles, is one of the famous problems in artificial intelligence. Along with chess, tic-tac-toe and backgammon, it has been used to study search algorithms.
The Eight Puzzle can be generalized into an M × N Puzzle where at least one of M and N is odd. The puzzle is constructed with MN − 1 sliding tiles with each a number from 1 to MN − 1 on it packed into a M by N frame with one tile missing. For example, with M = 4 and N = 3, a puzzle may look like:
| 1 | 6 | 2 |
| 4 | 0 | 3 |
| 7 | 5 | 9 |
| 10 | 8 | 11 |
Let's call missing tile 0. The only legal operation is to exchange 0 and the tile with which it shares an edge. The goal of the puzzle is to find a sequence of legal operations that makes it look like:
| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 10 | 11 | 0 |
The following steps solve the puzzle given above.
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START |
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DOWN |
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LEFT ⇒ |
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UP |
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… |
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RIGHT |
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UP |
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UP ⇒ |
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LEFT |
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GOAL |
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Given an M × N puzzle, you are to determine whether it can be solved.
Input
The input consists of multiple test cases. Each test case starts with a line containing M and N (2 ≤ M, N ≤ 999). This line is followed by M lines containing N numbers each describing an M × N puzzle.
The input ends with a pair of zeroes which should not be processed.
Output
Output one line for each test case containing a single word YES if the puzzle can be solved and NO otherwise.
Sample Input
3 3 1 0 3 4 2 5 7 8 6 4 3 1 2 5 4 6 9 11 8 10 3 7 0 0 0
Sample Output
YES NO
题意：给你一个m*n的矩阵，0代表空，0位置可以和上下左右位置交换，问是否可以变成1~m*n-1顺序排列且0在第m*n的位置的图，看上面例子。
思路：这就是一个奇数码问题的扩展，我们将其看成一条链，将0去除。
①对于n的奇数码问题，我们知道若能从一个图转换成另一张图，只需要比较两个图的逆序对奇偶性是否相同即可。(上下交换，交换n-1个数，n-1为偶数，不影响逆序对奇偶性)
②对于n的偶数码问题，我们左右交换依然不增减逆序对，但是上下交换，将交换n-1个数，n-1为奇数，将改变逆序对奇偶性，我们需要判断 【一个图的逆序对+空位置行的差值】与【另一图的逆序对】
1 #include<cstdio> 2 #include<iostream> 3 4 using namespace std; 5 6 const int maxn = 1e6+6; 7 int a[maxn]; 8 typedef long long ll; 9 int Mergesort(int l,int r) 10 { 11 int mid = (l+r)/2; 12 int b[r-l+5]; 13 int i=l,j=mid+1; 14 int m=0; 15 int cnt = 0; 16 while(i <= mid && j <= r) 17 { 18 if(a[i] > a[j]) 19 b[m++] = a[j++],cnt += mid-i+1; 20 else 21 b[m++] = a[i++]; 22 } 23 while(i <= mid) 24 { 25 b[m++] = a[i++]; 26 } 27 while(j <= r) 28 { 29 b[m++] = a[j++]; 30 } 31 m = 0; 32 for(int i=l; i<=r; i++) 33 { 34 a[i] = b[m++]; 35 } 36 return cnt; 37 } 38 39 void Merge(int l,int r,ll& ans) 40 { 41 if(l >= r) 42 return; 43 int mid = (l+r)/2; 44 Merge(l,mid,ans); 45 Merge(mid+1,r,ans); 46 ans += Mergesort(l,r); 47 } 48 int n,m; 49 int main() 50 { 51 while(~scanf("%d%d",&n,&m)&&n&&m) 52 { 53 int tmp; 54 int cnt = 1; 55 int row; 56 for(int i=1; i<=n; i++) 57 { 58 for(int j=1; j<=m; j++) 59 { 60 scanf("%d",&tmp); 61 if(tmp) 62 a[cnt++] = tmp; 63 else row = i; 64 } 65 } 66 ll ans = 0; 67 Merge(1,cnt-1,ans); 68 int flag = 0; 69 if(m&1){if((ans & 1) == 0)flag = 1;} 70 else if((ans+n-row) % 2 == 0)flag = 1; 71 if(flag)printf("YES "); 72 else printf("NO "); 73 } 74 }