题意

传送门 Codeforces 1579G Minimal Coverage

题解

DP

$dp[i+1][j]$ 代表 $0\cdots i$ 次移动后所在位置与覆盖区域最左侧位置相差 $j$ 时，覆盖区域的最小值。枚举左右方向递推即可。总时间复杂度 O ( n ⋅ max ⁡ { a i } ) O(n\cdot\max\{a_i\}) O(n⋅max{ai​})。

#include <bits/stdc++.h>
using namespace std;
constexpr int M = 2E3, INF = 1e9;int main() {ios::sync_with_stdio(false);cin.tie(nullptr);int tt;cin >> tt;while (tt--) {int n;cin >> n;vector<int> a(n);for (int i = 0; i < n; ++i) {cin >> a[i];}auto _min = [](int &x, int y) {x = min(x, y);};vector<vector<int>> dp(n + 1, vector<int>(M, INF));dp[0][0] = 0;for (int i = 0; i < n; ++i) {for (int j = 0; j < M; ++j) {if (dp[i][j] == INF) {continue;}if (j + a[i] < M) {_min(dp[i + 1][j + a[i]], max(dp[i][j], j + a[i]));}if (j - a[i] >= 0) {_min(dp[i + 1][j - a[i]], dp[i][j]);} else {_min(dp[i + 1][0], dp[i][j] + a[i] - j);}}}int res = *min_element(dp[n].begin(), dp[n].end());cout << res << '\n';}return 0;
}

二分 + bitset

二分覆盖区域的大小 $d$。用 std::bitset 维护当前的可能位置，初始位置可能位于 $[0,d)$ 中的任一个位置，递推即可。总时间复杂度 O ( n ⋅ max ⁡ { a i } ⋅ log ⁡ n / 32 ) O(n\cdot\max\{a_i\}\cdot\log n/32) O(n⋅max{ai​}⋅logn/32)。

#include <bits/stdc++.h>
using namespace std;
constexpr int N = 2E3;
using bt = bitset<N>;int main() {ios::sync_with_stdio(false);cin.tie(nullptr);int tt;cin >> tt;while (tt--) {int n;cin >> n;vector<int> a(n);for (int i = 0; i < n; ++i) {cin >> a[i];}auto judge = [&](int d) {bt x, mask;for (int i = 0; i < d; ++i) {x[i] = mask[i] = 1;}for (int i = 0; i < n; ++i) {x = ((x << a[i]) | (x >> a[i])) & mask;}return x.any();};int lb = 0, ub = N;while (ub - lb > 1) {int mid = (lb + ub) / 2;if (judge(mid)) {ub = mid;} else {lb = mid;}}cout << ub - 1 << '\n';}return 0;
}