ZOJ Problem Set - 2004 Commedia dell'arte

求解N*N*N多维棋盘问题，解题思路参考：再论八数码
求解逆序数参考：逆序数的几种求法
下面是C++源代码

#include<iostream>
#include<string.h>
using namespace std;#define N 1000005
int a[N];
int b[N];
int count;
long long inv;
//
void merge(int first, int mid, int last)
{int i, j, k;memcpy(b + first, a + first, (last - first + 1)*sizeof(int));k = i = first, j = mid + 1;while (i <= mid || j <= last){if (j > last || (i <= mid&&b[j] >= b[i])){a[k++] = b[i++];inv += (j - (mid + 1)) & 1;}else{a[k++] = b[j++];//inv += ((mid + 1) - i) & 1;}}
}void msort(int first,int last)
{if (first>=last)return;int mid = (first + last) / 2;msort(first, mid);msort(mid + 1, last);merge(first, mid, last);
}int main()
{int n,m,size,y,z,dis,s,v,i,j,iter;cin >> n;for (i = 0; i < n; i++){cin >> m;size = m * m * m;s = m * m;for (iter = j = 0; j < size; j++){cin >> v;v ? a[iter++] = v : (z = j / s, y = j % s / m);}inv = 0;msort(0,iter-1);if (m % 2 == 0){inv += (s - 1) * (m - 1 - z) + (m - 1) * (m - 1 - y);}/*if (m % 2 == 1){ans = inv % 2;}else{dis = (size - 1) / m - z / m;ans = (dis + inv) % 2;}*/if (!(inv&1))    cout << "Puzzle can be solved." << endl;else{cout << "Puzzle is unsolvable." << endl;}}//system("pause");
}

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