Statement
有\(n\)个节点, 分别用红线,蓝线连成两棵树. 用\(y\)种颜色给节点染色, 规定如果一条边在两棵树中同时出现, 那么边两端的点的颜色必须相同.
- Task #1: 给定两棵树, 求染色方案.
- Task #2: 给定其中一棵树, 求对于另一棵树的每一种形态的染色方案数之和.
- Task #3: 两棵树的形态都没有确定, 求对于所有情况的染色方案数之和.
(\(n\le 10^5\))
Solution
Task #1
计算有多少条公共边即可.
Task #2
如果有\(i\)条公共边, 那么贡献为\(y^{n-i}\). 设\(z=y^{-1}\), 贡献为\(y^n z^{i}\), 考虑计算\(z^i\),最后乘上\(y^n\).
\[
\begin{aligned}
z^i &= (z-1+1)^i \&= \sum_{j=0}^i (z-1)^j \binom{i}{j}
\end{aligned}
\]
于是我们对于边集\(E\)的每个子集\(E'\)的贡献为\((z-1)^{|E'|}\).
对于一个公共边的边集的选取方案, 将边连起来, 形成了很多个连通块, 大小为\(a_1\dots a_m\), 那么贡献为:
\[
(z-1)^{n-m} n^{m-2} \prod_{i=1}^m a_i
\]
\((z-1)^{n-m}\)为贡献, \(n^{m-2}\prod_{i=1}^m a_i\)为\(n\)个点,\(m\)个块的树的个数.
上面这个式子的组合意义为在每个块中选一个点的方案数, 树形DP即可, 时间复杂度\(O(n)\).
Task #3
类似Task2的思路, 枚举至少有\(n-m\)条公共边, 即\(m\)个连通块的方案数:
\[
\begin{aligned}
g_m
&=
\sum_{\sum\limits_{i=1}^m a_i = n}
\sum_{j=1}^m \binom{\sum_{k=1}^j a_k - 1}{a_j-1}
\left(n^{m-2} \prod_{i=1}^m a_i\right)^2
\prod _{i=1}^m a_i ^ {a_i - 2} \&=
\sum_{\sum\limits_{i=1}^m a_i = n}
\sum_{j=1}^m \binom{\sum_{k=1}^j a_k - 1}{a_j-1}
n^{2(m-2)}
\prod_{i=1}^m a_i^{a_i}
\end{aligned}
\]
\[
\begin{aligned}
ans &= \sum_m g_{m} \times (z - 1)^{n-m} \&=
\sum_{\sum\limits_{i=1}^m a_i = n}
\sum_{j=1}^m \binom{\sum_{k=1}^j a_k - 1}{a_j-1}
n^{2(m-2)}
\prod_{i=1}^m a_i^{a_i}
(z-1)^{n-m}\\end{aligned}
\]
设\(f_k\)表示当前已经处理了\(k\)个块, 那么有转移:
\[
f_k =
\begin{cases}
n^{-4}(z-1)^n
&,k=0 \(z-1)^{-1} n^2\sum_{i=1}^k f_{k-i} \binom{k-1}{i-1} i^i
&,k=1
\end{cases}
\]
写成卷积的形式:
\[
f_k = (z-1)^{-1}(k-1)!n^2 \sum_{i=1}^k \frac{f_{k-i}}{(k-i)!}\times\frac{i^i}{(i-1)!}
\]
分治FFT即可, 时间复杂度\(O(n\log^2n)\).
namespace Subtask0 {
map<LL, int> mp;
void Main() {
For(i, 1, n - 1) {
int x = read<int>(), y = read<int>();
mp[(LL)x * 1e9 + y] = 1;
}
int ans = 0;
For(i, 1, n - 1) {
int x = read<int>(), y = read<int>();
if (mp[(LL)x * 1e9 + y]) ++ ans;
}
printf("%d\n", fpm(y, n - ans));
}
}
namespace Subtask1 {
int beg[MAXN], v[MAXN << 1], nex[MAXN << 1], e = 1;
void add(int uu, int vv) {
v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e;
}
int dp[MAXN][2];
void DFS(int u, int pa) {
dp[u][0] = (LL)fpm(n, MOD - 2) * fpm(z - 1, n - 1) % MOD;
for (int i = beg[u]; i; i = nex[i])
if (v[i] != pa) {
DFS(v[i], u);
int t0 = dp[u][0], t1 = dp[u][1];
dp[u][0] = (LL)t0 * dp[v[i]][1] % MOD * n % MOD * n % MOD * fpm(fpm(z - 1), n) % MOD;
dp[u][1] = (LL)t1 * dp[v[i]][1] % MOD * n % MOD * n % MOD * fpm(fpm(z - 1), n) % MOD;
(dp[u][0] += (LL)t0 * dp[v[i]][0] % MOD * n % MOD * fpm(fpm(z - 1), n - 1) % MOD) %= MOD;
(dp[u][1] += (LL)t0 * dp[v[i]][1] % MOD * n % MOD * fpm(fpm(z - 1), n - 1) % MOD) %= MOD;
(dp[u][1] += (LL)t1 * dp[v[i]][0] % MOD * n % MOD * fpm(fpm(z - 1), n - 1) % MOD) %= MOD;
}
(dp[u][1] += dp[u][0]) %= MOD;
}
void Main() {
if (y == 1) {
printf("%d\n", fpm(n, n - 2));
return;
}
For(i, 2, n) {
int u = read<int>(), v = read<int>();
add(u, v), add(v, u);
}
DFS(1, 0);
printf("%lld\n", (LL)fpm(y, n) * dp[1][1] % MOD);
}
}
namespace Subtask2 {
int f[MAXN << 1];
void cdqFFT(int l, int r) {
static int A[MAXN << 1], B[MAXN << 1];
if (l == r) {
if (!l) f[0] = (LL)fpm(fpm(n), 4) * fpm(z_, n) % MOD;
else f[l] = (LL)f[l] * iz_ % MOD * fac[l - 1] % MOD;
return;
}
int mid = (l + r) >> 1, len = 1, pt = 0;
cdqFFT(l, mid);
while (len < mid - l + 1) ++ pt, len <<= 1;
Rep(i, len) {
A[i] = (LL)f[l + i] * ifac[l + i] % MOD;
if (i) B[i] = (LL)fpm(i, i) * ifac[i - 1] % MOD * n % MOD * n % MOD;
}
For(i, len, len * 2 - 1) {
A[i] = 0;
B[i] = (LL)fpm(i, i) * ifac[i - 1] % MOD * n % MOD * n % MOD;
}
calcrev(len << 1, pt + 1);
NTT(A, len << 1, 1), NTT(B, len << 1, 1);
Rep(i, len << 1) A[i] = (LL)A[i] * B[i] % MOD;
NTT(A, len << 1, -1);
For(i, mid + 1, r) inc(f[i], A[i - l]);
cdqFFT(mid + 1, r);
}
void Main() {
if (y == 1) {
printf("%d\n", fpm(n, 2 * n - 4));
return;
}
InitFac(n);
z_ = z - 1, iz_ = fpm(z_);
cdqFFT(0, N - 1);
printf("%lld\n", (LL)f[n] * fpm(y, n) % MOD);
}
}原文地址：https://www.cnblogs.com/Hany01/p/10357729.html