20250929
T1
制作美食
注意到只有 \(a\) 的前缀最大值是有用的，其他东西要么孤立要么和之前的前缀最大值连通。然后只需要对每个 \(b\) 找到连的是哪个区间，并查集合并一下即可。当然也可能孤立，就直接贡献答案。
代码
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
#define lowbit(x) ((x) & (-(x)))
#define getchar() p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++
char buf[1<<21], *p1, *p2, ch;
long long read() {
long long ret = 0, neg = 0; char c = getchar(); neg = (c == '-');
while (c < '0' || c > '9') c = getchar(), neg |= (c == '-');
while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
return ret * (neg ? -1 : 1);
}
int n, m;
int a[2000005], b[2000005];
struct BIT {
int bit[2000005];
void add(int x, int y) { for (; x <= n; x += lowbit(x)) bit[x] += y; }
int query(int x) {
int ret = 0;
for (; x; x -= lowbit(x)) ret += bit[x];
return ret;
}
} bit;
int o[2000005];
int p[2000005], pcnt;
int q[2000005], qcnt;
int ans;
int stk[2000005], sz;
int dsu[2000005];
int getf(int x) { return dsu[x] == x ? x : (dsu[x] = getf(dsu[x])); }
int main() {
freopen("delidish.in", "r", stdin);
freopen("delidish.out", "w", stdout);
int tc = read();
while (tc--) {
ans = pcnt = qcnt = sz = 0;
n = read(), m = read();
for (int i = 1; i <= n; i++) a[i] = read(), o[i] = i, dsu[i] = i, bit.bit[i] = 0;
for (int i = 1; i <= m; i++) b[i] = read();
sort(o + 1, o + n + 1, [](int x, int y) { return a[x] < a[y]; });
for (int i = 1, j = 1; i <= n; i++) {
while (j <= m && j <= a[o[i]]) bit.add(n - b[j] + 1, 1), ++j;
if (bit.query(n - o[i] + 1)) p[++pcnt] = o[i];
else ++ans;
}
sort(p + 1, p + pcnt + 1);
for (int i = 1, mx = 0; i <= pcnt; i++) if (a[p[i]] > mx) q[++qcnt] = p[i], mx = a[p[i]];
dsu[pcnt] = pcnt;
for (int i = 1, j = 1; i <= m; i++) {
while (j <= qcnt && i > a[q[j]]) ++j;
int t = upper_bound(q + 1, q + qcnt + 1, b[i]) - q - 1;
if (j <= t) for (int x = getf(j); x < t; x = getf(x)) dsu[x] = x + 1;
else ++ans;
}
for (int i = 1; i <= qcnt; i++) ans += (getf(i) == i);
cout << ans << "\n";
}
return 0;
}
T2
树与排列
\(f_{i, j}\) 表示 \(i\) 子树，分成了 \(j\) 个连续段的 \(\sum\prod\)。把子树拼起来之后可以钦定一些子树的连续段进行合并并乘上自己深度的贡献。但是不能合并相同子树的连续段。
-
因此对非法合并容斥。加入一棵子树时在子树里钦定一些非法合并然后乘容斥系数求和即可。这样每个出现了非法合并的方案的每个非法合并的子集都被乘了容斥系数求和，因此贡献为 \(0\)。而没有非法合并的贡献就是 \(1\)。那么做完了。
-
考虑将乘积拆开，即为在每个 LCA 的到根链上选一个点的方案数。那么发现这样我们就可以合并相同子树的连续段了，只是不必再乘上自己深度的贡献。那么直接做就好了。
代码（容斥）
#include <iostream>
#include <string.h>
#define int long long
using namespace std;
const int P = 1000000007;
inline void Madd(int &x, int y) { (x += y) >= P ? (x -= P) : 0; }
inline int Msum(int x, int y) { return Madd(x, y), x; }
int n;
int head[505], nxt[505], to[505], ecnt;
inline void add(int u, int v) { to[++ecnt] = v, nxt[ecnt] = head[u], head[u] = ecnt; }
int C[505][505];
int f[505][505], tmp[505];
int sz[505], fa[505];
int pw[505][505];
void dfs(int x, int d) {
f[x][sz[x] = 1] = 1;
for (int i = head[x]; i; i = nxt[i]) {
int v = to[i];
dfs(v, d + 1);
memset(tmp, 0, sizeof tmp);
for (int a = 1; a <= sz[v]; a++) {
for (int b = a + 1; b <= sz[v]; b++) {
if ((b - a) & 1) Madd(f[v][a], P - f[v][b] * C[b - 1][b - a] % P * pw[d][b - a] % P);
else Madd(f[v][a], f[v][b] * C[b - 1][b - a] % P * pw[d][b - a] % P);
}
}
for (int a = 1; a <= sz[x]; a++) {
for (int b = 1; b <= sz[v]; b++)
Madd(tmp[a + b], f[x][a] * f[v][b] % P * C[a + b][b] % P);
}
sz[x] += sz[v];
memcpy(f[x], tmp, sizeof tmp);
}
for (int i = 1; i <= sz[x]; i++) {
for (int j = i + 1; j <= sz[x]; j++)
Madd(f[x][i], f[x][j] * C[j - 1][j - i] % P * pw[d][j - i] % P);
}
}
signed main() {
freopen("tree.in", "r", stdin);
freopen("tree.out", "w", stdout);
cin >> n;
for (int i = C[0][0] = 1; i <= n; i++) {
for (int j = C[i][0] = 1; j <= i; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
for (int j = pw[i][0] = 1; j <= n; j++) pw[i][j] = pw[i][j - 1] * i % P;
}
for (int i = 2; i <= n; i++) cin >> fa[i], add(fa[i], i);
dfs(1, 1);
cout << f[1][1] << "\n";
return 0;
}
代码（product trick）
#include <iostream>
#include <string.h>
#define int long long
using namespace std;
const int P = 1000000007;
inline void Madd(int &x, int y) { (x += y) >= P ? (x -= P) : 0; }
inline int Msum(int x, int y) { return Madd(x, y), x; }
int n;
int head[505], nxt[505], to[505], ecnt;
inline void add(int u, int v) { to[++ecnt] = v, nxt[ecnt] = head[u], head[u] = ecnt; }
int C[505][505];
int f[505][505], tmp[505];
int sz[505], fa[505];
int pw[505][505];
void dfs(int x, int d) {
f[x][sz[x] = 1] = 1;
for (int i = head[x]; i; i = nxt[i]) {
int v = to[i];
dfs(v, d + 1);
memset(tmp, 0, sizeof tmp);
for (int a = 1; a <= sz[x]; a++) {
for (int b = 1; b <= sz[v]; b++)
Madd(tmp[a + b], f[x][a] * f[v][b] % P * C[a + b][b] % P);
}
sz[x] += sz[v];
memcpy(f[x], tmp, sizeof tmp);
}
for (int i = 1; i <= sz[x]; i++) {
for (int j = i + 1; j <= sz[x]; j++)
Madd(f[x][i], f[x][j] * C[j - 1][j - i] % P);
}
}
signed main() {
freopen("tree.in", "r", stdin);
freopen("tree.out", "w", stdout);
cin >> n;
for (int i = C[0][0] = 1; i <= n; i++) {
for (int j = C[i][0] = 1; j <= i; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
for (int j = pw[i][0] = 1; j <= n; j++) pw[i][j] = pw[i][j - 1] * i % P;
}
for (int i = 2; i <= n; i++) cin >> fa[i], add(fa[i], i);
dfs(1, 1);
cout << f[1][1] << "\n";
return 0;
}
T3
树上区域颜色并
考虑如何对子树数颜色，只需要对子树每个颜色维护出现的最浅深度。然后开一个线段树维护这个最浅深度的计数数组，以支持单点修改区间求和。那么离线做完了。在线考虑直接把 dsu on tree 对桶的所有修改可持久化下来就好了。由于强制在线不完全，可以在每个询问的时候知道我们到底想要哪些颜色的最浅深度，于是最浅深度的那个数组不需要可持久化。那么就做完了，空间 \(\mathcal{O}(m \log^2 m)\)，时间 \(\mathcal{O}(nq + q \log m + m \log^2 m)\)。
代码
#include <iostream>
#include <string.h>
#include <vector>
#include <array>
#include <set>
using namespace std;
#define getchar() p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++
char buf[1<<21], *p1, *p2, ch;
long long read() {
long long ret = 0, neg = 0; char c = getchar(); neg = (c == '-');
while (c < '0' || c > '9') c = getchar(), neg |= (c == '-');
while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
return ret * (neg ? -1 : 1);
}
int n, m, q;
vector<int> G1[200005], G2[200005];
int v1[25], v2[200005];
int rec[25];
array<int, 21> _clrcnt[1000005];
vector<int> vec[200005];
array<int, 5> qs[1000005];
int dfn[200005], _dfn[200005], ncnt;
int sz[200005], son[200005], fa[200005];
int dep[200005];
void dfs1(int x, int _f) {
dep[x] = dep[_f] + 1;
sz[x] = 1; fa[x] = _f;
_dfn[dfn[x] = ++ncnt] = x;
for (int v : G2[x]) {
if (v != _f) {
dfs1(v, x);
sz[x] += sz[v];
if (sz[v] > sz[son[x]]) son[x] = v;
}
}
}
int rt[200005];
struct node { int l, r, s; } T[86000005];
struct Segment_Tree {
int ncnt;
void Insert(int &p, int q, int l, int r, int x, int y) {
T[p = ++ncnt] = T[q]; T[p].s += y;
if (l == r) return;
int mid = (l + r) >> 1;
if (x <= mid) Insert(T[p].l, T[q].l, l, mid, x, y);
else Insert(T[p].r, T[q].r, mid + 1, r, x, y);
}
int Query(int o, int l, int r, int L, int R) {
if (!o) return 0;
if (L <= l && r <= R) return T[o].s;
int mid = (l + r) >> 1;
if (R <= mid) return Query(T[o].l, l, mid, L, R);
if (L > mid) return Query(T[o].r, mid + 1, r, L, R);
return Query(T[o].l, l, mid, L, R) + Query(T[o].r, mid + 1, r, L, R);
}
} seg;
multiset<int> st[200005];
int crt;
int mind[100005];
void _add(int x, int y) {
seg.Insert(crt, crt, 1, m + 1, mind[v2[x]], -1);
if (y == -1) st[v2[x]].erase(st[v2[x]].find(dep[x]));
else st[v2[x]].insert(dep[x]);
mind[v2[x]] = *st[v2[x]].begin(); // m + 1
seg.Insert(crt, crt, 1, m + 1, mind[v2[x]], 1);
}
void add(int x, int y) { for (int i = dfn[x]; i < dfn[x] + sz[x]; i++) _add(_dfn[i], y); }
void dfs2(int x) {
for (int v : G2[x]) if (v != fa[x] && v != son[x]) dfs2(v), add(v, -1);
if (son[x]) dfs2(son[x]);
for (int v : G2[x]) if (v != fa[x] && v != son[x]) add(v, 1);
_add(x, 1);
rt[x] = crt;
for (auto v : vec[x]) for (int i = 1; i <= n; i++) _clrcnt[v][i] = mind[_clrcnt[v][i]];
}
int _dep2[25], L[25], R[25], _ncnt;
void _dfs(int x, int fa) {
_dep2[x] = _dep2[fa] + 1; L[x] = ++_ncnt;
for (int v : G1[x]) if (v != fa) _dfs(v, x);
R[x] = _ncnt;
}
bool ap[100005];
int main() {
freopen("tree.in", "r", stdin);
freopen("tree.out", "w", stdout);
n = read(), m = read(), q = read();
for (int i = 1; i < n; i++) {
int u = read(), v = read();
G1[u].emplace_back(v);
G1[v].emplace_back(u);
}
for (int i = 1; i < m; i++) {
int u = read(), v = read();
G2[u].emplace_back(v);
G2[v].emplace_back(u);
}
for (int i = 1; i <= 100000; i++) st[i].insert(m + 1), mind[i] = m + 1;
for (int i = 1; i <= n; i++) v1[i] = read(), rec[i] = v1[i];
for (int i = 1; i <= m; i++) v2[i] = read();
for (int i = 1; i <= q; i++) {
qs[i][0] = read();
if (qs[i][0] == 1) {
qs[i][1] = read(), qs[i][2] = read(), qs[i][3] = read(), qs[i][4] = read();
for (int j = 1; j <= n; j++) _clrcnt[i][j] = v1[j];
vec[qs[i][2]].emplace_back(i);
} else {
qs[i][1] = read(), qs[i][2] = read();
v1[qs[i][1]] = qs[i][2];
}
}
dfs1(1, 0);
dfs2(1);
_dfs(1, 0);
memcpy(v1, rec, sizeof rec);
int lst = 0;
for (int i = 1; i <= q; i++) {
if (qs[i][0] == 1) {
int x = qs[i][1], y = qs[i][2], a = qs[i][3] ^ lst, b = qs[i][4] ^ lst;
int ans = seg.Query(rt[y], 1, m + 1, dep[y], min(m, dep[y] + b));
for (int j = 1; j <= n; j++) if (L[x] <= L[j] && L[j] <= R[x] && _dep2[j] - _dep2[x] <= a) {
if (_clrcnt[i][j] > min(m, dep[y] + b)) {
ans += !ap[v1[j]];
ap[v1[j]] = 1;
}
}
cout << (lst = ans) << "\n";
for (int j = 1; j <= n; j++) ap[v1[j]] = 0;
} else v1[qs[i][1]] = qs[i][2];
}
return 0;
}
T4
序排速快
由于 partition 之后递归再 bubble 和直接 bubble 没有区别，因此只需要考虑 bubble 的总长度。对于每个位置算贡献。对于位置 \(i\)，将 \(<i\) 的设为 \(0\)，\(i\) 设为 \(1\)，\(>i\) 的设为 \(2\)，那么如果 \(0\) 和 \(1\) 出现的最大位置为 \(p\)，则这个位置总共会贡献 \(p - i\) 次。当然如果 \(1\) 不在所有 \(0\) 的后面，则会有额外一次贡献来把 \(1\) 扔到最后。那么用所有贡献 \(p - i + 1\) 的方案减掉贡献 \(p - i\) 的方案，列出式子：
\(\sum\limits_{i = 1}^n\sum\limits_{j = i}^ni!(n - i)!\binom{j - 1}{i - 1}(j - i + 1) - \sum\limits_{i = 1}^n\sum\limits_{j = i}^n(i - 1)!(n - i)!\binom{j - 1}{i - 1}\)。前后分别处理：
\[\begin{equation}\begin{split}
\sum\limits_{i = 1}^n\sum\limits_{j = i}^n(i - 1)!(n - i)!\binom{j - 1}{i - 1} &=
\sum\limits_{i = 1}^n(i - 1)!(n - i)!\sum\limits_{j = i}^n\binom{j - 1}{i - 1} \\
&= \sum\limits_{i = 1}^n(i - 1)!(n - i)!\binom{n}{i} \\
&= \sum\limits_{i = 1}^nn!/i \\&= n!\sum\limits_{i = 1}^n\frac1{i}
\\
\sum\limits_{i = 1}^n\sum\limits_{j = i}^ni!(n - i)!\binom{j - 1}{i - 1}(j - i + 1)
&= \sum\limits_{i = 1}^ni!(n - i)!\sum\limits_{j = i}^n\binom{j - 1}{i - 1}j - \sum\limits_{j = i}^ni!(n - i)!\binom{j - 1}{i - 1}(i - 1) \\
&= \sum\limits_{i = 1}^ni!(n - i)!\sum\limits_{j = i}^n\binom{j - 1}{i - 1}j - i!(n - i)!(i - 1)\binom{n}{i} \\
&= \sum\limits_{i = 1}^ni!(n - i)!\sum\limits_{j = i}^n\binom{j - 1}{i - 1}j - n!(i - 1) \\
&= \sum\limits_{i = 1}^ni!(n - i)!(i\binom{n}{i} + (n - i + 1)\binom{n}{i} - \binom{n + 1}{i + 1}) - n!(i - 1) \\
&= \sum\limits_{i = 1}^n(n + 1)! - (n + 1)!/(i + 1) - n!(i - 1) \\
&= n(n + 1)! - \frac{n!n(n - 1)}{2} - (n + 1)!\sum\limits_{i = 1}^n\frac1{i + 1} \\
\end{split}\end{equation}
\]
那么预处理倒数和，即可 \(\mathcal{O}(1)\) 求出一个 \(n\) 的答案。那么就做完了。
代码
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
#define int long long
using namespace std;
const int P = 998244353, i2 = (P + 1) / 2;
int fac[10000005], pi[10000005], inv[10000005];
void pre(int n) {
fac[0] = fac[1] = pi[1] = inv[0] = inv[1] = 1;
for (int i = 2; i <= n; i++) {
fac[i] = fac[i - 1] * i % P;
inv[i] = (P - P / i) * inv[P % i] % P;
pi[i] = (pi[i - 1] + inv[i]) % P;
}
}
int L, R;
signed main() {
freopen("tros.in", "r", stdin);
freopen("tros.out", "w", stdout);
cin >> L >> R;
pre(R + 2);
int ans = 0;
for (int i = L; i <= R; i++) {
int tmp = (i * fac[i + 1] % P + P - fac[i] * i % P * (i - 1) % P * i2 % P + P - fac[i] * pi[i] % P + P - fac[i + 1] * (pi[i + 1] - 1) % P) % P;
ans ^= tmp;
}
cout << ans << "\n";
return 0;
}
T2，对非法限制容斥，product trick。
T3，转换贡献形式，对颜色算贡献。
T4，冒泡排序的刻画。
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