AtCoder Grand Contest 046 E Permutation Cover

洛谷传送门

AtCoder 传送门

若 \(2\min\limits_{i = 1}^m a_i < \max\limits_{i = 1}^n a_i\) 就无解，因为根据排列的性质必然存在 \(yxxxy\) 或两端 \(xxyy\) 的情况，并且若这个条件不满足，就可以构造一组解。

考虑最小化字典序。枚举排列和之前重合的长度，从小到大尝试加入。

如果 \(2mn + 1 < mx\) 就无解，如果 \(2mn \ge mx\) 就一定有解。

如果 \(2mn + 1 = mx\)，那么要先填 \(x\)，再填 \(y\)。

取字典序最小的情况即可。

code

// Problem: E - Permutation Cover
// Contest: AtCoder - AtCoder Grand Contest 046
// URL: https://atcoder.jp/contests/agc046/tasks/agc046_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 1010;

int m, a[maxn], n, b[maxn];
bool vis[maxn];
basic_string<int> ans, res;

inline void work(int x) {
	mems(vis, 0);
	for (int i = 1; i <= m - x; ++i) {
		vis[ans[(int)ans.size() - i]] = 1;
	}
	basic_string<int> vc;
	for (int i = 1; i <= m; ++i) {
		if (!vis[i]) {
			--b[i];
			vc += i;
		}
	}
	int mn = *min_element(b + 1, b + m + 1), mx = *max_element(b + 1, b + m + 1);
	if (mn * 2 + 1 < mx) {
		return;
	} else if (mn * 2 >= mx) {
		res = min(res, vc);
	} else {
		basic_string<int> S;
		int lst = 0;
		for (int i : vc) {
			if (b[i] == mx) {
				lst = i;
			}
		}
		for (int i = 0; i < (int)vc.size(); ++i) {
			int x = vc[i];
			if (b[x] != mn || x > lst) {
				S += x;
			}
			if (x == lst) {
				for (int j = 0; j < i; ++j) {
					if (b[vc[j]] == mn) {
						S += vc[j];
					}
				}
			}
		}
		res = min(res, S);
	}
}

void solve() {
	scanf("%d", &m);
	for (int i = 1; i <= m; ++i) {
		scanf("%d", &a[i]);
		n += a[i];
	}
	int mn = *min_element(a + 1, a + m + 1), mx = *max_element(a + 1, a + m + 1);
	if (mn * 2 < mx) {
		puts("-1");
		return;
	}
	while ((int)ans.size() < n) {
		res = basic_string<int>(1, 1e9);
		for (int i = ans.size() ? 1 : m; i <= min(n - (int)ans.size(), m); ++i) {
			for (int j = 1; j <= m; ++j) {
				b[j] = a[j];
			}
			work(i);
		}
		for (int i : res) {
			--a[i];
		}
		ans += res;
	}
	for (int x : ans) {
		printf("%d ", x);
	}
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}