CodeForces 1927G Paint Charges

洛谷传送门

CF 传送门

看到 $n \le 100$ 考虑 $O(\text{poly}(n))$ dp。发现从左向右决策，因为一个点可以向左或向右覆盖，所以需要记最靠左的未覆盖的位置 $j$ 和最靠右的已覆盖位置 $k$，状态就是 $f_{i, j, k}$，dp 最小的覆盖次数。

转移的讨论很简单。考虑不覆盖还是向左或向右覆盖，若向左或向右覆盖看能否覆盖到 $j$ 或 $k$。这样转移是 $O(1)$ 的。总时间复杂度 $O(n^3)$。

code

// Problem: G. Paint Charges
// Contest: Codeforces - Codeforces Round 923 (Div. 3)
// URL: https://codeforces.com/contest/1927/problem/G
// Memory Limit: 256 MB
// Time Limit: 4000 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 = 110;
 
int n, a[maxn], f[maxn][maxn][maxn];
 
void solve() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
	}
	mems(f, 0x3f);
	f[0][1][0] = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n + 1; ++j) {
			for (int k = 0; k <= n; ++k) {
				f[i][j][k] = min(f[i][j][k], f[i - 1][j][k]);
				if (max(i - a[i] + 1, 1) <= j && j <= i) {
					int nk = max(i, k);
					f[i][nk + 1][nk] = min(f[i][nk + 1][nk], f[i - 1][j][k] + 1);
				}
				int nk = min(i + a[i] - 1, n);
				if (nk > k) {
					int nj = (i <= j && j <= nk ? nk + 1 : j);
					f[i][nj][nk] = min(f[i][nj][nk], f[i - 1][j][k] + 1);
				}
			}
		}
	}
	printf("%d\n", f[n][n + 1][n]);
}
 
int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}