下面是一些 C++ 实现的算法,该算法基于递归的双射证明
!n = (n-1) * (!(n-1) + !(n-2)),
where !n
是混乱的数量n
items.
#include <algorithm>
#include <ctime>
#include <iostream>
#include <vector>
static const int N = 12;
static int count;
template<class RAI>
void derange(RAI p, RAI a, RAI b, int n) {
if (n < 2) {
if (n == 0) {
for (int i = 0; i < N; ++i) p[b[i]] = a[i];
if (false) {
for (int i = 0; i < N; ++i) std::cout << ' ' << p[i];
std::cout << '\n';
} else {
++count;
}
}
return;
}
for (int i = 0; i < n - 1; ++i) {
std::swap(a[i], a[n - 1]);
derange(p, a, b, n - 1);
std::swap(a[i], a[n - 1]);
int j = b[i];
b[i] = b[n - 2];
b[n - 2] = b[n - 1];
b[n - 1] = j;
std::swap(a[i], a[n - 2]);
derange(p, a, b, n - 2);
std::swap(a[i], a[n - 2]);
j = b[n - 1];
b[n - 1] = b[n - 2];
b[n - 2] = b[i];
b[i] = j;
}
}
int main() {
std::vector<int> p(N);
clock_t begin = clock();
std::vector<int> a(N);
std::vector<int> b(N);
for (int i = 0; i < N; ++i) a[i] = b[i] = i;
derange(p.begin(), a.begin(), b.begin(), N);
std::cout << count << " permutations in " << clock() - begin << " clocks for derange()\n";
count = 0;
begin = clock();
for (int i = 0; i < N; ++i) p[i] = i;
while (std::next_permutation(p.begin(), p.end())) {
for (int i = 0; i < N; ++i) {
if (p[i] == i) goto bad;
}
++count;
bad:
;
}
std::cout << count << " permutations in " << clock() - begin << " clocks for next_permutation()\n";
}
在我的机器上,我得到
176214841 permutations in 13741305 clocks for derange()
176214841 permutations in 14106430 clocks for next_permutation()
恕我直言,这是一种洗涤。可能双方都需要做出改进(例如,重新实施next_permutation
通过仅扫描发生变化的元素的混乱测试);这留给读者作为练习。