Problem T2. 作業

考慮狀壓 DP

\(f[mask]\) 表示當前已經排好序的顏色狀態為 \(mask\)，最少的交換次數。

預處理移動的最小代價即可。

時間複雜度 \(\mathcal O(400\times 2^{20}+20n\log n)\)

#include <bits/stdc++.h>

using namespace std;

const int N = 300005, M = 21;

int n;
vector<int> pos[M];
long long _move[M][M];
long long dp[1 << M];

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    pos[x].push_back(i);
  }
  for (int i = 1; i <= 20; ++i) {
    for (int j = 1; j <= 20; ++j) {
      if (i == j) 
        continue; 
      _move[i][j] = 0;
      for (int k = 0; k < (int)pos[i].size(); ++k) {
        if (pos[j].size() == 0 || pos[j][0] > pos[i][k])
          continue; 
        _move[i][j] += lower_bound(pos[j].begin(), pos[j].end(), pos[i][k]) - pos[j].begin(); 
      }
    }
  }
  memset(dp, 0x3f, sizeof dp); 
  dp[0] = 0; 
  for (int mask = 0; mask < (1 << 20); ++mask) {
    for (int j = 0; j < 20; ++j) {
      if (!(mask >> j & 1)) {
        long long res = 0;
        for (int k = 0; k < 20; ++k) {
          if (mask >> k & 1) {
            res += _move[k + 1][j + 1];
          }
        }
        dp[mask | (1 << j)] = min(dp[mask | (1 << j)], dp[mask] + res); 
      }
    }
  }
  cout << dp[(1 << 20) - 1] << '\n';
  return 0; 
}
 

Problem T3. Five-seventeen

將兩個路徑拆成向上和向下的路徑，考慮樹上倍增維護答案

為了減少邊界的處理，令 \(prep_{i,j}\) 表示從 \(i\) 的父親開始，向上倍增 \(j\) 次的答案

時間複雜度 \(\mathcal O(10\times n\log n)\)

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 1;

struct edge {
  int to, nxt;
} e[N << 1];

int head[N];
int edge_cnt;

void add_edge(int u, int v) {
  e[++edge_cnt] = (edge){v, head[u]};
  head[u] = edge_cnt;
}

const int LOG = 17; 

int n, m, q;
int bz[LOG + 1][N][11];
vector<int> tag[N];
int father[LOG + 1][N];
int dep[N];

void Dfs(int u, int Father) {
  father[0][u] = Father;
  dep[u] = dep[Father] + 1; 
  for (int i = 1; i <= LOG; ++i) 
    father[i][u] = father[i - 1][father[i - 1][u]]; 
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == Father)
      continue;
    Dfs(v, u); 
  }
}

int lca(int u, int v) {
  if (dep[u] < dep[v]) 
    swap(u, v);
  for (int i = LOG; i >= 0; --i) 
    if (dep[father[i][u]] >= dep[v]) 
      u = father[i][u];
  if (u == v) 
    return u;
  for (int i = LOG; i >= 0; --i) 
    if (father[i][u] != father[i][v]) 
      u = father[i][u], v = father[i][v];
  return father[0][u];
}

int c[11];

void merge(int* a, int* b) {
  int x = 1, y = 1;
  c[0] = 0;
  for (int i = 1; i <= 10; ++i) {
    if (x == a[0] + 1 && y == b[0] + 1)
      break;
    else if (x == a[0] + 1) 
      c[++c[0]] = b[y++];
    else if (y == b[0] + 1)
      c[++c[0]] = a[x++];
    else 
      c[++c[0]] = a[x] < b[y] ? a[x++] : b[y++];
  }
  for (int i = 0; i <= c[0]; ++i) 
    a[i] = c[i]; 
}

void merge(int* a, vector<int> b) {
  int x = 1, y = 0;
  c[0] = 0;
  for (int i = 1; i <= 10; ++i) {
    if (x == a[0] + 1 && y >= (int)b.size())
      break;
    else if (x == a[0] + 1) 
      c[++c[0]] = b[y++];
    else if (y >= (int)b.size())
      c[++c[0]] = a[x++];
    else 
      c[++c[0]] = a[x] < b[y] ? a[x++] : b[y++];
  }
  for (int i = 0; i <= c[0]; ++i) 
    a[i] = c[i]; 
}

void Dfs_1(int u) {
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == father[0][u]) 
      continue;
    Dfs_1(v);
    merge(bz[0][v], tag[u]); 
  }
}

void climb_up(int u, int v, int* ans) {
  for (int i = LOG; i >= 0; --i) {
    if (dep[father[i][u]] > dep[v]) {
      merge(ans, bz[i][u]); 
      u = father[i][u];
    }
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    add_edge(u, v);
    add_edge(v, u);
  }
  Dfs(1, 0);
  for (int i = 1; i <= n; ++i) 
    tag[i].clear(); 
  for (int i = 1; i <= m; ++i) {
    int x;
    cin >> x;
    tag[x].push_back(i); 
  }
  Dfs_1(1); 
  for (int i = 1; i <= n; ++i) {
    sort(bz[0][i] + 1, bz[0][i] + bz[0][i][0] + 1); 
    sort(tag[i].begin(), tag[i].end());
    tag[i].resize(min((int)tag[i].size(), 10)); 
  }
  for (int i = 1; i <= LOG; ++i) 
    for (int j = 1; j <= n; ++j) {
      for (int k = 0; k <= bz[i][j][0]; ++k) 
        bz[i][j][k] = bz[i - 1][j][k];
      merge(bz[i][j], bz[i - 1][father[i - 1][j]]); 
    }
  int ans[11];
  for (int qt = 1; qt <= q; ++qt) { 
    int u, v, k; 
    cin >> u >> v >> k; 
    int _lca = lca(u, v); 
    ans[0] = 0; 
    merge(ans, tag[u]);
    if (u != v) 
      merge(ans, tag[v]);
    if (u == _lca) {
      climb_up(v, u, ans);  
    } else if (v == _lca) {
      climb_up(u, v, ans);  
    } else {
      merge(ans, tag[_lca]); 
      climb_up(u, _lca, ans);
      climb_up(v, _lca, ans); 
    }
    cout << min(k, ans[0]) << ' ';
    for (int i = 1; i <= min(k, ans[0]); ++i) 
      cout << ans[i] << ' ';
    cout << '\n';
  }
  return 0; 
}