uva 716

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题目链接：uva 716 - Commedia dell' arte

题目大意：给定一个三维的八数码，0表示空的位置，问说是否可以排回有序序列。

解题思路：对于n为奇数的情况，考虑三维八数码对应以为状态下去除0的时候逆序对数，偶数的情况下，考虑将0的位置转移到（n，n，n）位置后对应序列的逆序对数。如果逆序对数为偶数即为可以，奇数不可以。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
int n, arr[maxn], t[maxn];
int x, y, z;
ll merge_sort (int l, int r, int* a, int* b) {
if (l == r)
return 0;
ll ret = 0;
int mid = (r + l) / 2;
int p = l, q = mid+1, mv = l;
ret = merge_sort(l, mid, a, b) + merge_sort(mid + 1, r, a, b);
while (p <= mid || q <= r) {
if (q > r || (p <= mid && a[p] < a[q]))
b[mv++] = a[p++];
else {
ret += mid - p + 1;
b[mv++] = a[q++];
}
}
for (int i = l; i <= r; i++)
a = b;
return ret;
}
bool judge () {
if (n&1) {
ll ret = merge_sort(0, n*n*n-1, arr, t) - (z * n * n + x * n + y - 1);
return (ret&1) == 0;
}
while (z != n - 1) {
int p = z * n * n + x * n + y;
z++;
int q = z * n * n + x * n + y;
swap(arr[p], arr[q]);
}
while (x != n - 1) {
int p = z * n * n + x * n + y;
x++;
int q = z * n * n + x * n + y;
swap(arr[p], arr[q]);
}
while (y != n - 1) {
int p = z * n * n + x * n + y;
y++;
int q = z * n * n + x * n + y;
swap(arr[p], arr[q]);
}
ll ret = merge_sort(0, n*n*n-2, arr, t);
return (ret&1) == 0;
}
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
scanf("%d", &n);
for (int k = 0; k < n; k++) {
int Z = k * n * n;
for (int i = 0; i < n; i++) {
int X = i * n;
for (int j = 0; j < n; j++) {
int tmp = Z + X + j;
scanf("%d", &arr[tmp]);
if (arr[tmp] == 0) {
x = i; y = j; z = k;
}
}
}
}
printf("%s\n", judge() ? "Puzzle can be solved." : "Puzzle is unsolvable.");
}
return 0;
}