10.10 总结

T1 美丽的子区间

还行吧，根据大眼观察法 可以看出当 \(x\) 为使用科技的次数时，函数 \(f(x)\) 等于使用 \(x\) 次科技的最小答案是一个单谷函数，可以三分，注意到使用 \(x\) 次科技的时候的第 \(i\) 个数的答案是 \(\min \limits_{j=\min(1,i-x+1)}^{i}\)。而且还要加上一个小贪心：把最小的放在最前面。

#include <fstream>
#include <cstring>

using namespace std;
using ll = long long;

const int kMaxN = 1e5 + 1;

ifstream cin("collect.in");
ofstream cout("collect.out");

ll n, k, o[kMaxN], a[kMaxN], s[kMaxN];

struct ST {
  static const int kMaxN = 2e5 + 5;
  ll n, log2[kMaxN], Maxn[kMaxN][20], Minn[kMaxN][20];
  ST() {}
  ST(ll n, ll a[]) {
    memset(Minn, 0x3f, sizeof(Minn));
    for (int i = 1; i <= n; i++) {
      Minn[i][0] = Maxn[i][0] = a[i];
    }
    for (int i = 2; i <= n; i++) {
      log2[i] = log2[i / 2] + 1;
    }
    for (int j = 1; j < 20; j++) {
      for (int i = 1; i + (1 << j - 1) <= n; i++) {
        Maxn[i][j] = max(Maxn[i][j - 1], Maxn[i + (1 << j - 1)][j - 1]);
        Minn[i][j] = min(Minn[i][j - 1], Minn[i + (1 << j - 1)][j - 1]);
      }
    }
  }
  ll Max(int l, int r) {
    int len = r - l + 1;
    return max(Maxn[l][log2[len]], Maxn[r - (1 << log2[len]) + 1][log2[len]]);
  }
  ll Min(int l, int r) {
    int len = r - l + 1;
    return min(Minn[l][log2[len]], Minn[r - (1 << log2[len]) + 1][log2[len]]);
  }
} st;

ll F(ll x) {
  ll ans = x * k;
  for (int i = 1; i <= n; i++) {
    ll o = st.Min(max(1ll, i - x), i);
    ans += o;
  }
  return ans;
}

int main() {
  cin >> n >> k;
  ll minn = 1e9, pos;
  for (int i = 1; i <= n; i++) {
    cin >> o[i];
    if (minn > o[i]) {
      pos = i;
      minn = o[i];
    }
  }
  for (int i = pos; i <= n; i++) {
    a[i - pos + 1] = o[i];
  }
  for (int i = 1; i < pos; i++) {
    a[i + (n - pos + 1)] = o[i];
  }
  st = ST(n, a);
  ll l = 0, r = n - 1;
  while (l < r) {
    ll mid1 = l + (r - l) / 3, mid2 = r - (r - l) / 3;
    ll ans1 = F(mid1), ans2 = F(mid2);
    if (ans1 > ans2) {
      l = mid1 + 1;
    } else {
      r = mid2 - 1;
    }
  }
  ll ans = 1e18;
  for (int i = max(0ll, l - 2); i <= min(n, r + 2); i++) {
    ans = min(ans, F(i));
  }
  cout << ans << '\n';
  return 0;
}