CF762F Tree nesting

题目连接

问题分析

可以给小树钦定一个根， \(Dp[i][j]\) 表示大树上的点 \(i\) 对应到小树上的点 \(j\) 的可能的方案数。然后每一步转移都是一个状压DP（将小树是否被匹配状压，然后枚举大树上的点和小树上的点匹配）。

但如果这样统计的话，在两种情况下有重复：

• 在小树取不同的根但仍同构；
• 确定小树的根后，小树的子树同构。

所以我们对钦定根后的小树进行哈希，即可排除第一种重复。而如果小树的某两个子树同构，那么就在统计的时候强行钦定一个顺序，这样就解决了第二种重复。

参考程序

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>

const int Maxn = 2000;
const int Maxm = 12;
const int MaxAlpha = 1 << Maxm;
int Mod = 1e9 + 7;
struct edge
{
    int To, Next;
    edge() {}
    edge(int _To, int _Next) : To(_To), Next(_Next) {}
};
struct node {
    int Value, Index;
    node() {}
    node(int _Value, int _Index) : Value(_Value), Index(_Index) {}
    inline bool operator<(const node Other) const
    {
        return Value < Other.Value;
    }
};
edge Edge1[Maxn << 1], Edge2[Maxm << 1];
int n, m, Ans;
int Start1[Maxn + 1], Start2[Maxm + 1], Used1, Used2;
int Hash[Maxm + 1], Father[Maxm + 1], Size[Maxm + 1];
int Dp[Maxn + 1][Maxm + 1], F[2][MaxAlpha];
node Temp[Maxm + 1]; int Ctrl[Maxm + 1];
std::set<int> Set;

inline void AddEdge1(int x, int y);
inline void AddEdge2(int x, int y);
inline void Init();
void GetHash(int u, int Fa);
void Calc(int u, int Fa);

int main()
{
    Init();
    for (int i = 1; i <= m; ++i)
    {
        GetHash(i, 0);
        if (Set.count(Hash[i]))
            continue;
        Set.insert(Hash[i]);
        memset(Dp, 0, sizeof(Dp));
        Calc(1, 0);
        for (int j = 1; j <= n; ++j)
            Ans = (Ans + Dp[j][i]) % Mod;
    }
    printf("%d\n", Ans);
    return 0;
}

inline void AddEdge1(int x, int y)
{
    Edge1[++Used1] = edge(y, Start1[x]);
    Start1[x] = Used1;
    return;
}

inline void AddEdge2(int x, int y)
{
    Edge2[++Used2] = edge(y, Start2[x]);
    Start2[x] = Used2;
    return;
}

inline void Init()
{
    scanf("%d", &n);
    for (int i = 1; i < n; ++i)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        AddEdge1(x, y);
        AddEdge1(y, x);
    }
    scanf("%d", &m);
    for (int i = 1; i < m; ++i)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        AddEdge2(x, y);
        AddEdge2(y, x);
    }
    return;
}

void GetHash(int u, int Fa)
{
    Size[u] = 1;
    Father[u] = Fa;
    for (int t = Start2[u]; t; t = Edge2[t].Next)
    {
        int v = Edge2[t].To;
        if (v == Fa)
            continue;
        GetHash(v, u);
        Size[u] += Size[v];
    }
    int Count = 0;
    for (int t = Start2[u]; t; t = Edge2[t].Next)
    {
        int v = Edge2[t].To;
        if (v == Fa)
            continue;
        Temp[++Count] = node(Hash[v], v);
    }
    std::sort(Temp + 1, Temp + Count + 1);
    Hash[u] = 0;
    for (int i = 1; i <= Count; ++i)
    {
        Hash[u] <<= Size[Temp[i].Index] << 1;
        Hash[u] += Hash[Temp[i].Index];
    }
    Hash[u] <<= 1;
    Hash[u] += 1 << ((Size[u] << 1) - 1);
    return;
}

void Calc(int u, int Fa)
{
    for (int t = Start1[u]; t; t = Edge1[t].Next)
    {
        int v = Edge1[t].To;
        if (v == Fa)
            continue;
        Calc(v, u);
    }
    for (int uu = 1; uu <= m; ++uu)
    {
        if (Size[uu] == 1)
        {
            Dp[u][uu] = 1;
            continue;
        }
        int Count = 0;
        memset(Ctrl, 0, sizeof(Ctrl));
        for (int t = Start2[uu]; t; t = Edge2[t].Next)
        {
            int v = Edge2[t].To;
            if (v ==Father[uu]) continue;
            Temp[++Count] = node(Hash[v], v);
        }
        std::sort(Temp + 1, Temp + Count + 1);
        for (int i = 2; i <= Count; ++i)
            if (Temp[i].Value == Temp[i - 1].Value)
                Ctrl[i] = 1;
        memset(F, 0, sizeof(F));
        F[0][0] = 1;
        int Step = 0;
        for (int t = Start1[u]; t; t = Edge1[t].Next)
        {
            int v = Edge1[t].To;
            if (v ==Fa)
                continue;
            for (int j = 0; j < 1 << Count; ++j)
                F[(Step + 1) & 1][j] = 0;
            for (int j = 0; j < 1 << Count; ++j)
                for (int k = 1; k <= Count; ++k)
                {
                    if ((j >> (k - 1)) & 1)
                        continue;
                    if (Ctrl[k] && ((j >> (k - 2)) & 1) == 0)
                        continue;
                    F[(Step + 1) & 1][j | (1 << (k - 1))] += 1LL * F[Step & 1][j] * Dp[v][Temp[k].Index] % Mod;
                    F[(Step + 1) & 1][j | (1 << (k - 1))] %= Mod;
                }
            for (int j = 0; j < 1 << Count; ++j)
            {
                F[(Step + 1) & 1][j] += F[Step & 1][j];
                F[(Step + 1) & 1][j] %= Mod;
            }
            ++Step;
        }
        Dp[u][uu] = F[Step & 1][(1 << Count) - 1];
    }
    return;
}