【题目链接】
【思路要点】
- 将树拆成若干路径,并考虑长度 ≥ 2 \geq2 ≥2 的路径的放置方向。
- 然后问题变成了在环上放点,相邻位置颜色不同的方案数。
- 做法同 【校内训练2018-06-28】比谁数得对 。
- 时间复杂度 O ( ( ∑ k i ) 2 ) O((\sum k_i)^2) O((∑ki)2) 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 5005; const int P = 998244353; template <typename T> void read(T &x) { x = 0; int f = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -f; for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; x *= f; } vector <int> a[MAXN]; int dp[MAXN][MAXN][3], size[MAXN], coef[MAXN][MAXN], val[MAXN][MAXN]; int N, s[MAXN], finalans[MAXN], fac[MAXN], inv[MAXN]; 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; } void update(int &x, int y) { x += y; if (x >= P) x -= P; } void work(int pos, int fa) { size[pos] = 1; dp[pos][1][0] = 1; for (unsigned i = 0; i < a[pos].size(); i++) if (a[pos][i] != fa) { int x = a[pos][i]; work(x, pos); static int res[MAXN][3]; for (int i = 0; i <= size[pos] + size[x]; i++) res[i][0] = res[i][1] = res[i][2] = 0; for (int i = 0; i <= size[pos]; i++) for (int j = 0; j <= size[x]; j++) { update(res[i + j][0], 1ll * dp[pos][i][0] * dp[x][j][0] % P); if (i + j >= 1) update(res[i + j - 1][1], 1ll * dp[pos][i][0] * dp[x][j][0] % P); update(res[i + j][0], 2ll * dp[pos][i][0] * dp[x][j][1] % P); if (i + j >= 1) update(res[i + j - 1][1], 1ll * dp[pos][i][0] * dp[x][j][1] % P); update(res[i + j][0], 2ll * dp[pos][i][0] * dp[x][j][2] % P); update(res[i + j][1], 1ll * dp[pos][i][1] * dp[x][j][0] % P); if (i + j >= 1) update(res[i + j - 1][2], 1ll * dp[pos][i][1] * dp[x][j][0] % P); update(res[i + j][1], 2ll * dp[pos][i][1] * dp[x][j][1] % P); if (i + j >= 1) update(res[i + j - 1][2], 1ll * dp[pos][i][1] * dp[x][j][1] % P); update(res[i + j][1], 2ll * dp[pos][i][1] * dp[x][j][2] % P); update(res[i + j][2], 1ll * dp[pos][i][2] * dp[x][j][0] % P); update(res[i + j][2], 2ll * dp[pos][i][2] * dp[x][j][1] % P); update(res[i + j][2], 2ll * dp[pos][i][2] * dp[x][j][2] % P); } for (int i = 0; i <= size[pos] + size[x]; i++) memcpy(dp[pos][i], res[i], sizeof(res[i])); size[pos] += size[x]; } } void init(int n) { fac[0] = inv[0] = 1; for (int i = 1; i <= n; i++) { fac[i] = 1ll * fac[i - 1] * i % P; inv[i] = power(fac[i], P - 2); } coef[1][1] = 1; for (int i = 2; i <= n; i++) for (int j = 1; j <= i; j++) { coef[i][j] = coef[i - 1][j - 1] - coef[i - 1][j]; if (coef[i][j] < 0) coef[i][j] += P; } } int inverse(int x) { return 1ll * inv[x] * fac[x - 1] % P; } int main() { int T; read(T), init(MAXN - 1); for (int t = 1; t <= T; t++) { int n; read(n); s[t] = n; N += n; for (int i = 1; i <= n; i++) { a[i].clear(); memset(dp[i], 0, sizeof(dp[i])); } for (int i = 1; i <= n - 1; i++) { int x, y; read(x), read(y); a[x].push_back(y); a[y].push_back(x); } work(1, 0); for (int i = 1; i <= n; i++) { int ways = (dp[1][i][0] + 2ll * dp[1][i][1] + 2ll * dp[1][i][2]) % P * fac[i] % P; if (t == 1) ways = 1ll * ways * inverse(i) % P; for (int j = 1; j <= i; j++) update(val[t][j], 1ll * ways * coef[i][j] % P); } } if (T == 1) { cout << 0 << endl; return 0; } finalans[0] = 1; int tot = 0; for (int t = 1; t <= T; t++) { static int tmp[MAXN]; for (int i = 0; i <= tot + s[t]; i++) tmp[i] = 0; if (t == 1) { for (int i = 0; i <= tot; i++) for (int j = 1; j <= s[t]; j++) update(tmp[i + j], 1ll * finalans[i] * val[t][j] % P * inv[j - 1] % P); } else { for (int i = 0; i <= tot; i++) for (int j = 1; j <= s[t]; j++) update(tmp[i + j], 1ll * finalans[i] * val[t][j] % P * inv[j] % P); } for (int i = 0; i <= tot + s[t]; i++) finalans[i] = tmp[i]; tot += s[t]; } int ans = 0; for (int i = 1; i <= tot; i++) update(ans, 1ll * finalans[i] * fac[i - 1] % P); memset(finalans, 0, sizeof(finalans)), finalans[0] = 1; tot = 0; for (int t = 1; t <= T; t++) { static int tmp[MAXN]; for (int i = 0; i <= tot + s[t]; i++) tmp[i] = 0; if (t == 1) { for (int i = 0; i <= tot; i++) for (int j = 2; j <= s[t]; j++) update(tmp[i + j], 1ll * finalans[i] * val[t][j] % P * inv[j - 2] % P); } else { for (int i = 0; i <= tot; i++) for (int j = 1; j <= s[t]; j++) update(tmp[i + j], 1ll * finalans[i] * val[t][j] % P * inv[j] % P); } for (int i = 0; i <= tot + s[t]; i++) finalans[i] = tmp[i]; tot += s[t]; } for (int i = 2; i <= tot; i++) update(ans, P - 1ll * finalans[i] * fac[i - 2] % P); cout << ans << endl; return 0; }