20240911

Jordan's Castles

我們先思考如何快速求出 \(b_1, b_2, b_3...b_n\) 顯然我們可以直接用二分找到,然後我們可以直接將 \(a_i\) 改為 \(min(a[i], b[i])\),然後統計答案即可

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e5 + 5;

int t, n, a[N];

void Solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int ans = 0;
  for (int i = n; i >= 1; i--) {
    int l = 0, r = n;
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (a[mid] >= i) {
        l = mid;
      }
      else r = mid - 1;
    }
    ans += max(0ll, a[i] - l);
    a[i] = min(a[i], l);
  }
  cout << ans << "\n";
}

signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}

Game on a Graph

我們看到一個重要的性質 "移除該分量中編號最小的頂點",那麼我們可以從大到小列舉編號,每次列舉出邊 \(v\) 那麼只要 \(v < i\) 就可以直接用並查集合並,最後判斷一下奇偶性即可

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e5 + 5;

int t, n, m, fa[N], f[N], sz[N];

vector<int> g[N], new_g[N];

void dfs(int u) {
  sz[u] = 1;
  for (auto v : new_g[u]) {
    dfs(v);
    sz[u] += sz[v];
  }
}

int find(int x) {
  if (fa[x] == x) {
    return x;
  }
  return fa[x] = find(fa[x]);
}

void Solve() {
  cin >> n >> m;
  for (int i = 0; i <= n; i++) {
    fa[i] = i;
    g[i].clear();
    new_g[i].clear();
  }
  for (int i = 1, u, v; i <= m; i++) {
    cin >> u >> v;
    u++, v++;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = n; i >= 1; i--) {
    for (auto v : g[i]) {
      if (v > i && find(v) != i) {
        f[find(v)] = i;
        new_g[i].push_back(find(v));
        fa[find(v)] = i;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (find(i) == i) {
      f[i] = 0;
      new_g[0].push_back(i);
    }
  }
  dfs(0);
  for (int i = 1; i <= n; i++) {
    if (f[i] == 0 || (n - sz[f[i]]) % 2 == 1) {
      cout << i - 1 << ' ';
    }
  }
  cout << "\n";
}

signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    Solve();
  }
  return 0;
}