POJ 2893 M × N Puzzle（树状数组求逆序对）

                                                           M × N Puzzle

Time Limit: 4000MS		Memory Limit: 131072K
Total Submissions: 4112		Accepted: 1140

Description

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：8数码问题的升级，就是通过移动空格（用0代替）使得原来状态变成有序的1234......0，不过，这题是N*M数码。

题解：N*M都挺大的，搜索必然不行。考虑终态，实际就是逆序数为0的状态，然后四种操作方式分为：左右移动，对原序列的逆序数不影响；上下移动，如下：

-------------0***********

***********x-------------

x是任意数，现在要把x移上去，那么***********中，假设有a个大于x，b个小于x，那么移动之后逆序数就会加上一个b-a，x所能影响的也就是这些罢了，除此之外，其他都不变。

接着，如果列数为偶数，那么******的个数就是奇数，b,a奇偶性互异，b-a为奇数，所以移动一次后，原序列的逆序数的奇偶性变了。

考虑到最后0会移动到最后一行，所以奇偶性会改变n-i次(i为0的行数)，只需判断最后是否是偶数即可。

反之，如果列数为奇数，那么******的个数就是偶数，b,a奇偶性相同，b-a为偶数，所以移动一次后，原序列的逆序数的奇偶性没变。

因为无论怎么移，奇偶性都不变，所以说一开始初态的奇偶性就必须与末态一致。

题意来自http://blog.csdn.net/tmeteorj/article/details/8530105

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 2e9
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = 999*999+5;
const int M = 4e5+5;
int n,m,tot=0,cnt=0;
int head[N],ans[N];
int tree[N];
int a[N];
void add(int k,int num) {
    while(k<=999*999-1) {
        tree[k]+=num;
        k+=k&(-k);
    }
}
int Sum(int k) {
    int sum=0;
    while(k>0) {
        sum+=tree[k];
        k-=k&(-k);
    }
    return sum;
}
int main() {

    while(scanf("%d%d",&n,&m),n||m) {
        int x,y,t,s=0,nu=0;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++) {
                scanf("%d",&t);
                if(t==0)
                    x=i,y=j;
                else
                    a[nu++]=t;
            }
       met(tree,0);
        for(int i=nu-1; i>=0; i--) {
            s+=Sum(a[i]-1);
            add(a[i],1);
        }
        if(m&1)
            if(s&1)puts("NO");
            else   puts("YES");
        else if(((n-x)^s)&1) puts("NO");
        else            puts("YES");
    }
    return 0;
}