【LOJ2320】「清華集訓 2017」生成樹計數
阿新 • • 發佈：2019-02-01
【題目連結】
【思路要點】
- 連上\(a_i\)的限制，題目要求的實際上是\(\sum_{T}\prod_{i=1}^{N}a_i^{d_i}*d_i^{M}\sum_{i=1}^{N}d_i^{M}\)。
- 我們知道樹的Prufer序列與樹點的度數密切相關，因此考慮使用Prufer序列解題。
- Prufer序列提供了一個長度為\(N-2\)，各個元素在\(N\)以內的正整數序列和\(N\)個節點的有標號無根樹的雙射，每個元素在Prufer序列中出現的次數+1就是其在樹上的點度。
- 考慮給定式子的組合意義：寫出一個Prufer序列，在序列後補足1到\(N\)各一個，使得各數出現次數等於其樹上的點度，在每一種數中選擇一個位置塗色（每次塗的顏色不同，且不會覆蓋），重複\(M\)次；再選取一個數\(X\)，重複對某個寫有\(X\)的位置塗色\(M\)次，最後將方案數乘以這個數列上各個數權值（\(a_i\)）的乘積，將結果加入答案。
- 記\(f_{i,j}\)表示考慮了1到\(i\)，尚未選取過\(X\)，Prufer序列中的\(N-2\)個位置有\(j\)個位置已經塗了色，可行的方案數與已經考慮過的數的權值乘積的和；\(g_{i,j}\)表示考慮了1到\(i\)，已經選取過\(X\)，Prufer序列中的\(N-2\)個位置有\(j\)個位置已經塗了色，可行的方案數與已經考慮過的數的權值乘積的和。
- 轉移時列舉對於\(i\)，有多少Prufer序列中的位置被塗了色，則有$$f_{i,j}=\sum_{k=0}^{M}f_{i-1,j-k}*\binom{N-2-j+k}{k}*a_i^{k+1}*(S(M,k)*k!+S(M,k+1)*(k+1)!)$$ $$g_{i,j}=\sum_{k=0}^{2*M}f_{i-1,j-k}*\binom{N-2-j+k}{k}*a_i^{k+1}*(S(2*M,k)*k!+S(2*M,k+1)*(k+1)!)$$ $$g_{i,j}=g_{i,j}+\sum_{k=0}^{M}g_{i-1,j-k}*\binom{N-2-j+k}{k}*a_i^{k+1}*(S(M,k)*k!+S(M,k+1)*(k+1)!)$$
- 其中\(S(M,k)\)表示第二類斯特林數。
- 最終計算答案有\(Ans=\sum_{i=0}^{N-2}g_{N,i}*(\sum_{j=1}^{N}a_j)^{N-2-i}\)。
- 直接實現上述DP，時間複雜度\(O(N^2M)\)。
- 上述DP的式子可以的化為卷積的形式：$$F_i=F_{i-1}*A_i,G_i=G_{i-1}*A_i+F_{i-1}*B_i$$
- 其中\(A_i,B_i\)為轉移係數多項式，\(A_i\)的次數為\(M\)，\(B_i\)的次數為\(2*M\)，我們希望求出\(G_N\)係數的最後\(N-1\)項。
- 分治NTT，將係數卷積起來，並每一步對\(x^N\)取模即可。
- 時間複雜度\(O(NMLogN)\)。
【程式碼】
- 暴力DP，時間複雜度\(O(N^2M)\)。
#include<bits/stdc++.h> using namespace std; const int MAXM = 65; const int MAXN = 6005; const int P = 998244353; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int power(int x, int y) { if (y == 0) return 1; int tmp = power(x, y / 2); if (y % 2 == 0) return 1ll * tmp * tmp % P; else return 1ll * tmp * tmp % P * x % P; } int n, m, a[MAXN], inv[MAXN], fac[MAXN]; int f[MAXN][MAXN], g[MAXN][MAXN], s[MAXM][MAXM], p[MAXN][MAXM]; int c(int x, int y) { return 1ll * fac[x] * inv[y] % P * inv[x - y] % P; } void update(int &x, int y) { x = (x + y) % P; } int main() { fac[0] = 1; for (int i = 1; i < MAXN; i++) fac[i] = 1ll * fac[i - 1] * i % P; inv[MAXN - 1] = power(fac[MAXN - 1], P - 2); for (int i = MAXN - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1ll) % P; s[0][0] = 1; for (int i = 1; i < MAXM; i++) for (int j = 1; j <= i; j++) s[i][j] = (s[i - 1][j - 1] + 1ll * j * s[i - 1][j]) % P; for (int i = 1; i < MAXM; i++) for (int j = 1; j <= i; j++) s[i][j] = (1ll * s[i][j] * fac[j]) % P; read(n), read(m); int sum = 0; for (int i = 1; i <= n; i++) { read(a[i]), sum = (sum + a[i]) % P; p[i][0] = 1; for (int j = 1; j < MAXM; j++) p[i][j] = 1ll * p[i][j - 1] * a[i] % P; } f[0][0] = 1; for (int i = 1; i <= n; i++) for (int j = 0; j <= n - 2; j++) { for (int k = 0; k <= j && k <= m; k++) { int tmp = 1ll * c(n - 2 - j + k, k) * p[i][k + 1] % P * s[m][k] % P; update(f[i][j], 1ll * f[i - 1][j - k] * tmp % P); if (k != m) { int tmp = 1ll * c(n - 2 - j + k, k) * p[i][k + 1] % P * s[m][k + 1] % P; update(f[i][j], 1ll * f[i - 1][j - k] * tmp % P); } } for (int k = 0; k <= j && k <= 2 * m; k++) { int tmp = 1ll * c(n - 2 - j + k, k) * p[i][k + 1] % P * s[2 * m][k] % P; update(g[i][j], 1ll * f[i - 1][j - k] * tmp % P); if (k != 2 * m) { int tmp = 1ll * c(n - 2 - j + k, k) * p[i][k + 1] % P * s[2 * m][k + 1] % P; update(g[i][j], 1ll * f[i - 1][j - k] * tmp % P); } } for (int k = 0; k <= j && k <= m; k++) { int tmp = 1ll * c(n - 2 - j + k, k) * p[i][k + 1] % P * s[m][k] % P; update(g[i][j], 1ll * g[i - 1][j - k] * tmp % P); if (k != m) { int tmp = 1ll * c(n - 2 - j + k, k) * p[i][k + 1] % P * s[m][k + 1] % P; update(g[i][j], 1ll * g[i - 1][j - k] * tmp % P); } } } int ans = 0; for (int i = 0; i <= n - 2; i++) update(ans, 1ll * g[n][i] * power(sum, n - 2 - i) % P); writeln(ans); return 0; }
- NTT優化後的DP，時間複雜度\(O(NMLogN)\)。
#include<bits/stdc++.h> using namespace std; const int MAXM = 65; const int MAXN = 262144; const int MAXLOG = 20; const int P = 998244353; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int inv[MAXN], fac[MAXN], s[MAXM][MAXM]; int n, m, a[MAXN], p[MAXN][MAXM]; int f[MAXLOG][MAXN], g[MAXLOG][MAXN]; namespace NTT { const int MAXN = 262144; const int P = 998244353; const int G = 3; int power(int x, int y) { if (y == 0) return 1; int tmp = power(x, y / 2); if (y % 2 == 0) return 1ll * tmp * tmp % P; else return 1ll * tmp * tmp % P * x % P; } int N, Log, home[MAXN]; void NTTinit() { for (int i = 0; i < N; i++) { int ans = 0, tmp = i; for (int j = 1; j <= Log; j++) { ans <<= 1; ans += tmp & 1; tmp >>= 1; } home[i] = ans; } } void NTT(int *a, int mode) { for (int i = 0; i < N; i++) if (home[i] < i) swap(a[i], a[home[i]]); for (int len = 2; len <= N; len <<= 1) { int delta; if (mode == 1) delta = power(G, (P - 1) / len); else delta = power(G, P - 1 - (P - 1) / len); for (int i = 0; i < N; i += len) { int now = 1; for (int j = i, k = i + len / 2; k < i + len; j++, k++) { int tmp = a[j]; int tnp = 1ll * a[k] * now % P; a[j] = (tmp + tnp) % P; a[k] = (tmp - tnp + P) % P; now = 1ll * now * delta % P; } } } if (mode == -1) { int inv = power(N, P - 2); for (int i = 0; i < N; i++) a[i] = 1ll * a[i] * inv % P; } } void times(int *fx, int *fy, int *gx, int *gy, int limit) { N = 1, Log = 0; while (N <= limit) { N <<= 1; Log++; } for (int i = limit + 1; i < N; i++) fx[i] = fy[i] = gx[i] = gy[i] = 0; NTTinit(); NTT(fx, 1), NTT(fy, 1); NTT(gx, 1), NTT(gy, 1); for (int i = 0; i < N; i++) { gx[i] = (1ll * fx[i] * gy[i] + 1ll * fy[i] * gx[i]) % P; fx[i] = 1ll * fx[i] * fy[i] % P; } NTT(fx, -1), NTT(gx, -1); for (int i = limit + 1; i < N; i++) fx[i] = gx[i] = 0; } } void update(int &x, int y) { x += y; if (x >= P) x -= P; } void init() { fac[0] = 1; for (int i = 1; i < MAXN; i++) fac[i] = 1ll * fac[i - 1] * i % P; inv[MAXN - 1] = NTT :: power(fac[MAXN - 1], P - 2); for (int i = MAXN - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1ll) % P; s[0][0] = 1; for (int i = 1; i < MAXM; i++) for (int j = 1; j <= i; j++) s[i][j] = (s[i - 1][j - 1] + 1ll * j * s[i - 1][j]) % P; for (int i = 1; i < MAXM; i++) for (int j = 1; j <= i; j++) s[i][j] = (1ll * s[i][j] * fac[j]) % P; } void solve(int l, int r, int depth) { int limit = min((r - l + 2) * m, n); for (int i = 0; i <= limit; i++) f[depth][i] = g[depth][i] = 0; if (l == r) { for (int k = 0; k <= m; k++) { update(f[depth][k], 1ll * inv[k] * p[l][k + 1] % P * s[m][k] % P); if (k != m) update(f[depth][k], 1ll * inv[k] * p[l][k + 1] % P * s[m][k + 1] % P); } for (int k = 0; k <= 2 * m; k++) { update(g[depth][k], 1ll * inv[k] * p[l][k + 1] % P * s[2 * m][k] % P); if (k != 2 * m) update(g[depth][k], 1ll * inv[k] * p[l][k + 1] % P * s[2 * m][k + 1] % P); } for (int i = limit + 1; i <= 2 * (limit + m); i++) f[depth][i] = g[depth][i] = 0; return; } int mid = (l + r) / 2; solve(l, mid, depth); solve(mid + 1, r, depth + 1); NTT :: times(f[depth], f[depth + 1], g[depth], g[depth + 1], min((mid - l + 2) * m, n) + min((r - mid + 1) * m, n)); for (int i = limit + 1; i <= 2 * (limit + m); i++) f[depth][i] = g[depth][i] = 0; } int main() { init(); read(n), read(m); int sum = 0; for (int i = 1; i <= n; i++) { read(a[i]), sum = (sum + a[i]) % P; p[i][0] = 1; for (int j = 1; j < MAXM; j++) p[i][j] = 1ll * p[i][j - 1] * a[i] % P; } solve(1, n, 0); int ans = 0; for (int i = 0; i <= n - 2; i++) update(ans, 1ll * g[0][i] * NTT :: power(sum, n - 2 - i) % P * fac[n - 2] % P * inv[n - 2 - i] % P); writeln(ans); return 0; }