标签：eof whether queue this gen amp start from proc
| 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.
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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：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; }
POJ 2893 M × N Puzzle（树状数组求逆序对）
标签：eof whether queue this gen amp start from proc
原文地址：http://www.cnblogs.com/jianrenfang/p/6091623.html
