loj 3217 「PA 2019」Desant - 动态规划 - 复杂度分析

#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;

const int N = 41;
const int inf = (signed) (~0u >> 2);

#define pii pair<int, long long>

pii operator + (pii a, pii b) {
	if (a.first == b.first)
		return pii(a.first, a.second + b.second);
	return min(a, b);
}

int n;
int p[N];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", p + i);
		--p[i];
	}
	vector<boolean> have (n, true);
	vector<int> lenc (n + 1, 1), lenp;
	vector<int> prodc (n + 2, 1), prodp;
	vector<pii> f {pii(0, 1)}, g;
	vector<int> vs;
	for (int i = 1; i <= n; i++) {
		swap(lenp, lenc);
		swap(prodp, prodc);
		swap(f, g);
		int id = 0;
		for (int j = 0; j < p[i]; j++)
			id += have[j];
		have[p[i]] = false;
		lenc.clear();
		prodc = {1};
		int ls = -1;
		for (int j = 0; j < n; j++) {
			if (have[j]) {
				lenc.push_back(j - ls);
				prodc.push_back(prodc.back() * lenc.back());
				ls = j;
			}
		}
		lenc.push_back(n - ls);
		prodc.push_back(prodc.back() * lenc.back());
		f = vector<pii>(prodc.back(), pii(inf, 0));
		for (int s = 0; s < prodp.back(); s++) {
			if (g[s].first == inf)
				continue;
			vs.clear();
			int t = s;
			for (auto l : lenp)
				vs.push_back(t % l), t /= l;
			int dlt = 0;
			for (int j = id + 1; j < (signed) vs.size(); j++)
				dlt += vs[j];
			vs[id] += vs[id + 1];
//			cerr << id << " " << vs.size() << ‘\n‘;
			vs.erase(vs.begin() + id + 1);
			int ns = 0;
			for (int j = 0; j < (signed) lenc.size(); j++)
				ns += vs[j] * prodc[j];
			f[ns] = f[ns] + g[s];
			g[s].first += dlt;
			ns += prodc[id];
			f[ns] = f[ns] + g[s];
		}
	}
	vector<pii> ans (n + 1, pii(inf, 0)); 
	for (int s = 0; s < prodc.back(); s++) {
		if (f[s].first == inf)
			continue;
		vs.clear();
		int t = s, len = 0;
		for (auto l : lenc)
			len += t % l, t /= l;
		ans[len] = ans[len] + f[s];
	}
	for (int i = 1; i <= n; i++) {
		printf("%d %lld\n", ans[i].first, ans[i].second);
	}
	return 0;
}