sgu 139 Help Needed!

题目描述：

139. Help Needed!

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

Little Johnny likespuzzles a lot! Just a few days ago, he found out about the 'traditional'4x4puzzle. For this puzzle, you have all the numbers from0 to 15arranged in 4 rows and 4 columns. You are allowed to switchtwo adjacent elements (horizontally or vertically), only if one of themhas the value0. The purpose of the puzzle is to reach the followingfinal state:

                            1  2  3  4
                            5  6  7  8
                            9 10 11 12
                           13 14 15  0

Given the initial stateof the puzzle, you have to decide whether there exists a sequence of moveswhich brings the puzzle into the final state.

Input

The input will consistof  4 lines, each of them containing 4 integers, describingthe initial state of the puzzle.

Output

For every initialstate, you should print "YES" if the final state can be reachedafter several moves or "NO", if such a thing is impossible.

Sample Input #1

1 2 3 45 6 7 89 10 11 013 14 15 12

Sample Output #1

YES

Sample Input #2

2 1 3 45 6 7 89 10 11 120 13 14 15

Sample Output #2

NO

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十五数码问题，手贱百度了一下，然后就找到结论，一套就AC了。

逆序对于0到底的哈密顿距离之和的奇偶性不变是好证的，至于为什么一定有解就难了。

贴个代码（重要的思维啊。。。我等爇菜啊。。。。）

#include<iostream>#include<cstring>#include<cstdio>#include<set>#include<algorithm>#include<vector>#include<cstdlib>#define inf 0xfffffff#define CLR(a,b) memset((a),(b),sizeof((a)))#define FOR(a,b) for(int a=1;a<=(b);(a)++)using namespace std;int const nMax = 1010;int const base = 10;typedef int LL;typedef pair<LL,LL> pij;//    std::ios::sync_with_stdio(false);int a[17];int main(){    int n(0),m(0);    FOR(i,16){cin>>a[i];if(a[i]==0)m=i;}    FOR(i,16) for(int j=i+1;j<=16;j++)if(a[j]<a[i])n++;    m=8-((m-1)/4+1+(m-1)%4+1);//cout<<m<<" "<<n<<endl;    m+=n;    if(m&1)printf("YES\n");    else printf("NO\n");    return 0;}