树 题解
一、题目：
二、思路：
先来简要介绍一下部分分。
第一组20分，dfs爆搜。
第二组20分，是一个菊花图。（在这里顺便提一句，在《组合数学》这本书上，菊花图被称作“星”，非常得有意思。）很显然我们需要分类讨论：
- 根节点不是关键点。答案：\(\sum\limits_{i=1}^m S(K,i)\)。
- 根节点是关键点。答案：\(\sum\limits_{i=1}^{m}S(K-1,i-1)\)。
其中，\(S\) 是第二类斯特林数。其实上面这两个和式都是贝尔数。
第三组60分。我们需要认真考虑一下。首先需要建出虚树来，在虚树上做树形DP。
设DP状态 \(dp[i,x,j]\) 为考虑了 \(x\) 的前 \(i\) 个儿子，分了 \(j\) 组的方案数。于是有 DP方程
\[dp[i,x,j]=\sum\limits_{j_1,j_2}dp[i-1,x,j_1]\times dp[y,j_2]\times\binom{j_1}{j_1+j_2-j}\times\binom{j_2}{j_1+j_2-j}\times(j_1+j_2-j)!
\]
\(y\) 是 \(x\) 的第 \(i\) 个儿子。
时间复杂度 \(O(n\times m^3)\)。但肯定跑不满。
第四组满分。我们发现这道题很像第二类斯特林数的模型。那么我们就需要考虑一下如何将斯特林数的模型运用到这道题上。
设 \(g_x\) 为 \(x\) 到根的路径中有多少个关键点。如果我们把关键点按照 \(g\) 排序，那么得到的序列就满足一个性质：\(x\) 的所有祖先全部排在 \(x\) 的前面。于是类比第二类斯特林数，我们设 \(S'(i,j)\) 为考虑了前 \(i\) 个关键点，分成不标号的 \(j\) 组的方案数。于是有
\[S'(i,j)=S'(i-1,j-1)+S'(i-1,j)\times \max\{j - g_x,0\}
\]
其中，\(x\) 是第 \(i\) 个关键点。
时间复杂度 \(O(nm)\)。
三、代码：
60分
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define FILEIN(s) freopen(s".in", "r", stdin);
#define FILEOUT(s) freopen(s".out", "w", stdout)
#define mem(s, v) memset(s, v, sizeof s)
inline int read(void) {
int x = 0, f = 1; char ch = getchar();
while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return f * x;
}
const int maxn = 1e4 + 5, logmaxn = 15;
const int maxm = 55, mod = 1e9 + 7;
int n, Q;
int dfn[maxn], num, fa[maxn][logmaxn], dep[maxn], K, m;
int query[maxn], stk[maxn], top, cnt[maxn];
long long dp[maxn][maxm], tmp[maxm], factor[maxm], C[maxm][maxm];
bool tag[maxn];
struct Graph {
int head[maxn], tot;
struct Edge {
int y, next;
Edge() {}
Edge(int _y, int _next) : y(_y), next(_next) {}
}e[maxn << 1];
inline void connect(int x, int y) {
e[++ tot] = Edge(y, head[x]);
head[x] = tot;
}
}A, B;
inline bool cmp(const int &a, const int &b) {
return dfn[a] < dfn[b];
}
void prework(int x, int father) {
fa[x][0] = father;
for (int i = 1; i <= 14; ++ i)
fa[x][i] = fa[fa[x][i - 1]][i - 1];
dep[x] = dep[father] + 1;
dfn[x] = ++ num;
for (int i = A.head[x]; i; i = A.e[i].next) {
int y = A.e[i].y;
if (y == father) continue;
prework(y, x);
}
}
void init(void) {
factor[0] = 1;
for (int i = 1; i <= 50; ++ i)
factor[i] = factor[i - 1] * i % mod;
C[0][0] = 1;
for (int i = 1; i <= 50; ++ i) {
C[i][0] = 1;
for (int j = 1; j <= i; ++ j)
C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
}
}
inline int LCA(int x, int y) {
if (dep[x] < dep[y]) swap(x, y);
for (int i = 14; i >= 0; -- i)
if (dep[fa[x][i]] >= dep[y])
x = fa[x][i];
if (x == y) return x;
for (int i = 14; i >= 0; -- i)
if (fa[x][i] != fa[y][i])
x = fa[x][i], y = fa[y][i];
return fa[x][0];
}
void insert(int x) {
if (x == 1) return;
if (top <= 1) { stk[++ top] = x; return; }
int lca = LCA(stk[top], x);
if (lca == stk[top]) { stk[++ top] = x; return; }
while (top > 1 && dfn[lca] <= dfn[stk[top - 1]]) {
B.connect(stk[top - 1], stk[top]);
B.connect(stk[top], stk[top - 1]);
-- top;
}
if (lca != stk[top]) {
B.connect(lca, stk[top]);
B.connect(stk[top], lca);
stk[top] = lca;
}
stk[++ top] = x;
}
void dfs(int x, int father) {
for (int i = B.head[x]; i; i = B.e[i].next) {
int y = B.e[i].y;
if (y == father) continue;
dfs(y, x);
}
dp[x][0] = 1;
for (int i = B.head[x]; i; i = B.e[i].next) {
int y = B.e[i].y;
if (y == father) continue;
swap(tmp, dp[x]);
mem(dp[x], 0);
for (int j = 0; j <= min(m, cnt[x] + cnt[y]); ++ j)
for (int j1 = 0; j1 <= min(j, min(m, cnt[x])); ++ j1)
for (int j2 = j - j1; j2 <= min(j, min(m, cnt[y])); ++ j2)
(dp[x][j] += tmp[j1] * dp[y][j2] % mod * C[j1][j1 + j2 - j] % mod * C[j2][j1 + j2 - j] % mod * factor[j1 + j2 - j] % mod) %= mod;
cnt[x] += cnt[y];
}
if (tag[x]) {
++ cnt[x];
for (int i = min(cnt[x], m); i >= 1; -- i) dp[x][i] = dp[x][i - 1];
dp[x][0] = 0;
}
}
void clear(int x, int father) {
for (int i = B.head[x]; i; i = B.e[i].next) {
int y = B.e[i].y;
if (y == father) continue;
clear(y, x);
}
tag[x] = false;
cnt[x] = 0;
for (int i = 1; i <= m; ++ i) dp[x][i] = 0;
B.head[x] = 0;
}
int main() {
FILEIN("tree"); FILEOUT("tree");
n = read(); Q = read();
for (int i = 1; i < n; ++ i) {
int x = read(), y = read();
A.connect(x, y); A.connect(y, x);
}
prework(1, 0);
init();
while (Q --) {
K = read(); m = read();
int r = read();
for (int i = 1; i <= K; ++ i)
tag[query[i] = read()] = true;
if (!tag[r]) query[++ K] = r;
sort(query + 1, query + K + 1, cmp);
stk[++ top] = 1;
for (int i = 1; i <= K; ++ i)
insert(query[i]);
while (top > 1) {
B.connect(stk[top - 1], stk[top]);
B.connect(stk[top], stk[top - 1]);
-- top;
}
dfs(r, 0);
long long ans = 0;
for (int i = 0; i <= m; ++ i) (ans += dp[r][i]) %= mod;
printf("%lld\n", ans);
top = 0; B.tot = 0;
clear(r, 0);
}
return 0;
}
100分
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define FILEIN(s) freopen(s".in", "r", stdin);
#define FILEOUT(s) freopen(s".out", "w", stdout)
#define mem(s, v) memset(s, v, sizeof s)
inline int read(void) {
int x = 0, f = 1; char ch = getchar();
while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return f * x;
}
const int maxn = 1e5 + 5, maxm = 305, logmaxn = 20, mod = 1e9 + 7;
int dfn[maxn], fa[maxn][logmaxn], dep[maxn], num;
int stk[maxn], top, n, Q, K, m, query[maxn], g[maxn], tmp[maxn];
long long S[maxn][maxm];
bool tag[maxn];
struct Graph {
int head[maxn], tot;
struct Edge {
int y, next;
Edge() {}
Edge(int _y, int _next) : y(_y), next(_next) {}
}e[maxn << 1];
inline void connect(int x, int y) {
e[++ tot] = Edge(y, head[x]);
head[x] = tot;
}
}A, B;
inline bool cmp(const int &a, const int &b) {
return dfn[a] < dfn[b];
}
inline bool cmp2(const int &a, const int &b) {
return g[a] < g[b];
}
void prework(int x, int father) {
fa[x][0] = father;
for (int i = 1; i <= 17; ++ i)
fa[x][i] = fa[fa[x][i - 1]][i - 1];
dep[x] = dep[father] + 1;
dfn[x] = ++ num;
for (int i = A.head[x]; i; i = A.e[i].next) {
int y = A.e[i].y;
if (y == father) continue;
prework(y, x);
}
}
inline int LCA(int x, int y) {
if (dep[x] < dep[y]) swap(x, y);
for (int i = 17; i >= 0; -- i)
if (dep[fa[x][i]] >= dep[y])
x = fa[x][i];
if (x == y) return x;
for (int i = 17; i >= 0; -- i)
if (fa[x][i] != fa[y][i])
x = fa[x][i], y = fa[y][i];
return fa[x][0];
}
void insert(int x) {
if (x == 1) return;
if (top <= 1) { stk[++ top] = x; return; }
int lca = LCA(stk[top], x);
if (lca == stk[top]) { stk[++ top] = x; return; }
while (top > 1 && dfn[lca] <= dfn[stk[top - 1]]) {
B.connect(stk[top - 1], stk[top]);
B.connect(stk[top], stk[top - 1]);
-- top;
}
if (lca != stk[top]) {
B.connect(lca, stk[top]);
B.connect(stk[top], lca);
stk[top] = lca;
}
stk[++ top] = x;
}
void dfs(int x, int father) {
g[x] = g[father] + tag[father];
for (int i = B.head[x]; i; i = B.e[i].next) {
int y = B.e[i].y;
if (y == father) continue;
dfs(y, x);
}
}
void clear(int x, int father) {
for (int i = B.head[x]; i; i = B.e[i].next) {
int y = B.e[i].y;
if (y == father) continue;
clear(y, x);
}
tag[x] = g[x] = B.head[x] = 0;
}
int main() {
FILEIN("tree"); FILEOUT("tree");
n = read(); Q = read();
for (int i = 1; i < n; ++ i) {
int x = read(), y = read();
A.connect(x, y); A.connect(y, x);
}
prework(1, 0);
while (Q --) {
K = read(); m = read();
int r = read();
for (int i = 1; i <= K; ++ i)
tag[tmp[i] = query[i] = read()] = true;
if (!tag[r]) tmp[++ K] = r;
sort(tmp + 1, tmp + K + 1, cmp);
stk[++ top] = 1;
for (int i = 1; i <= K; ++ i)
insert(tmp[i]);
while (top > 1) {
B.connect(stk[top - 1], stk[top]);
B.connect(stk[top], stk[top - 1]);
-- top;
}
dfs(r, 0);
if (!tag[r]) -- K;
sort(query + 1, query + K + 1, cmp2);
S[0][0] = 1;
for (int i = 1; i <= K; ++ i)
for (int j = 1; j <= min(m, i); ++ j)
S[i][j] = (S[i - 1][j - 1] + S[i - 1][j] * max(0, j - g[query[i]]) % mod) % mod;
long long ans = 0;
for (int i = 0; i <= m; ++ i)
(ans += S[K][i]) %= mod;
printf("%lld\n", ans);
B.tot = 0;
top = 0;
clear(r, 0);
}
return 0;
}
