M × N Puzzle POJ - 2893（奇数码）

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.

START

1	6	2
4	0	3
7	5	9
10	8	11

DOWN
⇒

1	0	2
4	6	3
7	5	9
10	8	11

LEFT
⇒

1	2	0
4	6	3
7	5	9
10	8	11

UP
⇒

1	2	3
4	6	0
7	5	9
10	8	11

…

RIGHT
⇒

1	2	3
4	0	6
7	5	9
10	8	11

UP
⇒

1	2	3
4	5	6
7	0	9
10	8	11

UP
⇒

1	2	3
4	5	6
7	8	9
10	0	11

LEFT
⇒

1	2	3
4	5	6
7	8	9
10	11	0

GOAL

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】

#include<cstdio>
#include<iostream>

using namespace std;

const int maxn = 1e6+6;
int a[maxn];
typedef long long ll;
int Mergesort(int l,int r)
{
    int mid = (l+r)/2;
    int b[r-l+5];
    int i=l,j=mid+1;
    int m=0;
    int cnt = 0;
    while(i <= mid && j <= r)
    {
        if(a[i] > a[j])
            b[m++] = a[j++],cnt += mid-i+1;
        else
            b[m++] = a[i++];
    }
    while(i <= mid)
    {
        b[m++] = a[i++];
    }
    while(j <= r)
    {
        b[m++] = a[j++];
    }
    m = 0;
    for(int i=l; i<=r; i++)
    {
        a[i] = b[m++];
    }
    return cnt;
}

void Merge(int l,int r,ll& ans)
{
    if(l >= r)
        return;
    int mid = (l+r)/2;
    Merge(l,mid,ans);
    Merge(mid+1,r,ans);
    ans += Mergesort(l,r);
}
int n,m;
int main()
{
    while(~scanf("%d%d",&n,&m)&&n&&m)
    {
        int tmp;
        int cnt = 1;
        int row;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                scanf("%d",&tmp);
                if(tmp)
                    a[cnt++] = tmp;
                else row = i;
            }
        }
        ll ans = 0;
        Merge(1,cnt-1,ans);
        int flag = 0;
        if(m&1){if((ans & 1) == 0)flag = 1;}
        else if((ans+n-row) % 2 == 0)flag = 1;
        if(flag)printf("YES\n");
        else printf("NO\n");
    }
}