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：

1. 根节点不是关键点。答案：(sumlimits_{i=1}^m S(K,i))。
2. 根节点是关键点。答案：(sumlimits_{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]=sumlimits_{j_1,j_2}dp[i-1,x,j_1] imes dp[y,j_2] imesinom{j_1}{j_1+j_2-j} imesinom{j_2}{j_1+j_2-j} imes(j_1+j_2-j)! ]

(y) 是 (x) 的第 (i) 个儿子。

时间复杂度 (O(n imes 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) imes 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
", 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
", ans);

        B.tot = 0;
        top = 0;
        clear(r, 0);
    }
    return 0;
}