codeforces1111 簡單題【DE】簡要題解

D

很顯然能夠用一個揹包算出來湊齊i個位置的方案

而後總的答案就是\(dp_{n / 2}\)

而後須要扣掉不符合條件的就是把選出來的數的貢獻剪掉的貢獻

而後注意由於是多重集合的排列，因此須要乘上\(\frac{fac[n / 2]}{fac[cnt_a]fac[cnt_b].....}\ast \frac{fac[n / 2]}{fac[cnt_c]fac[cnt_d].....}\)

而後顯然下面是全部個數的階乘積，而後沒了

#include<bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;
const int M = 60;
const int Mod = 1e9 + 7;

int add(int a, int b) {
  return (a += b) >= Mod ? a - Mod : a;
}

int sub(int a, int b) {
  return (a -= b) < 0 ? a + Mod : a;
}

int mul(int a, int b) {
  return 1ll * a * b % Mod;
}

int fast_pow(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) res = mul(res, a);
  return res;
}

int inv[N], fac[N];
int ans[M][M] = {0}, cnt[M];
int n, q, f[N];
char s[N];

int id(char c) {
  if ('a' <= c && c <= 'z') {
    return c - 'a' + 27;
  } else {
    return c - 'A' + 1;
  }
}

int C(int a, int b) {
  return a >= b ? mul(fac[a], mul(inv[b], inv[a - b])) : 0;
}

void modify(int id, int typ) {
  if (typ) {
    for (int i = n / 2; i >= cnt[id]; i--)
      f[i] = add(f[i], f[i - cnt[id]]);
  } else {
    for (int i = cnt[id]; i <= n / 2; i++)
      f[i] = sub(f[i], f[i - cnt[id]]);
  }
}

int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  scanf("%s", s + 1);
  n = strlen(s + 1);
  inv[0] = fac[0] = 1;
  for (int i = 1; i <= n; i++)
    fac[i] = mul(fac[i - 1], i);
  inv[n] = fast_pow(fac[n], Mod - 2);
  for (int i = n - 1; i >= 1; i--)
    inv[i] = mul(inv[i + 1], i + 1);
  for (int i = 1; i <= n; i++)
    cnt[id(s[i])]++;
  f[0] = 1;
  for (int i = 1; i <= 52; i++)
    if (cnt[i]) modify(i, 1);
  for (int i = 1; i <= 52; i++) if (cnt[i] && cnt[i] <= n / 2) {
    modify(i, 0);
    ans[i][i] = f[n / 2 - cnt[i]];
    modify(i, 1);
  }
  for (int i = 1; i <= 52; i++) if (cnt[i]) {
    modify(i, 0);
    for (int j = 1; j <= 52; j++) if (i != j && cnt[j] && cnt[j] + cnt[i] <= n / 2) {
      modify(j, 0);
      ans[i][j] = f[n / 2 - cnt[i] - cnt[j]];
      modify(j, 1);
    }
    modify(i, 1);
  }
  int bas = mul(fac[n / 2], fac[n / 2]);
  for (int i = 1; i <= 52; i++)
    bas = mul(bas, inv[cnt[i]]);
  for (int i = 1; i <= 52; i++)
    for (int j = 1; j <= 52; j++)
      ans[i][j] = mul(ans[i][j], mul(2, bas));
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    printf("%d\n", ans[id(s[x])][id(s[y])]);
  }
  return 0;
}

E

首先若是在虛樹上考慮，咱們按照深度進行dp

發現\(f_{i,j}\)表示i個點分紅j個集合的方案數有轉移：

\(f_{i,j}=f_{i - 1,j}\ast (j - h_i)+f_{i - 1,j - 1}\)

其中h是一個節點的父親個數

而後h咋算呢？

就是能夠用i到1的節點個數加上r到1的節點個數減去lca到1的節點個數的兩倍

而後以1爲根的時候每個點在dfs序上的貢獻都是一個區間，用bit算一下就能夠了

而後就直接dp就能夠了

由於有屢次詢問，因此注意邊界就能夠了

#include<bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;
const int M = 5e2 + 10;
const int LOG = 20;
const int Mod = 1e9 + 7;

int add(int a, int b) {
  return (a += b) >= Mod ? a - Mod : a;
}

int mul(int a, int b) {
  return 1ll * a * b % Mod;
}

int n, q, f[N][M];
vector<int> g[N];
int dep[N], fa[N][LOG];
int bg[N], ed[N], ind = 0;

void dfs(int u, int father) {
  dep[u] = dep[father] + 1;
  bg[u] = ++ind;
  fa[u][0] = father;
  for (int i = 1; i < 18; i++)
    fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (auto v : g[u])
    if (v != father)
      dfs(v, u);
  ed[u] = ind;
}

int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  int delta = dep[x] - dep[y];
  for (int i = 0; i < 18; i++)
    if ((delta >> i) & 1)
      x = fa[x][i];
  if (x == y) return x;
  for (int k = 17; k >= 0; k--) {
    if (fa[x][k] != fa[y][k]) {
      x = fa[x][k];
      y = fa[y][k];
    }
  }
  return fa[x][0];
}

int bit[N];

void modify(int t, int vl) {
  for (; t <= n; t += t & (-t))
    bit[t] += vl;
}

int query(int t) {
  int res = 0;
  for (; t; t -= t & (-t))
    res += bit[t];
  return res;
}

void solve() {
  static int h[N], p[N], k, m, r;
  static bool mark[N];
  scanf("%d %d %d", &k, &m, &r);
  for (int i = 1; i <= k; i++)
    scanf("%d", &p[i]);
  for (int i = 1; i <= k; i++) {
    modify(bg[p[i]], 1);
    modify(ed[p[i]] + 1, -1);
    mark[p[i]] = 1;
  }
  int hrt = query(bg[r]);
  for (int i = 1; i <= k; i++) {
    int g = lca(p[i], r);
    h[i] = query(bg[p[i]]) + hrt - 2 * query(bg[g]) + mark[g] - 1; 
  }
  for (int i = 1; i <= k; i++) {
    modify(bg[p[i]], -1);
    modify(ed[p[i]] + 1, 1);
    mark[p[i]] = 0;
  }
  sort(h + 1, h + k + 1); // 至關於在虛樹上按照深度進行dp
  f[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j < h[i]; j++)
      f[i][j] = 0;
    for (int j = h[i]; j <= min(i, m); j++)
      f[i][j] = add(mul(j - h[i], f[i - 1][j]), f[i - 1][j - 1]);
  }
  int res = 0;
  for (int i = 1; i <= m; i++)
    res = add(res, f[k][i]);
  printf("%d\n", res);
}

int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  scanf("%d %d", &n, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  while (q--)
    solve();
  return 0;
}