「JSOI2015」isomorphism

传送门

我们还是考虑树哈希来判同构。

但是我们需要使用一些特殊的手段来特殊对待假节点。

由于是无向树，我们首先求出重心，然后以重心为根跑树哈希。

此处我们不计算假节点的个数对子树大小的贡献。需要注意的是无向树可能有两个重心。

树哈希的时候，假节点儿子的哈希值也直接向上贡献（因为假节点有且只有一个儿子）。

这样我们就可以求出一颗无向树的简化树的哈希值，之后的问题就轻松解决了。

参考代码：

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}
 
typedef unsigned long long ull;
const int _ = 1e4 + 5;
const ull base = 19491001;
 
int tot, head[_]; struct Edge { int v, nxt; } edge[_ << 1];
inline void Add_edge(int u, int v) { edge[++tot] = (Edge) { v, head[u] }, head[u] = tot; }
 
int m, s[_]; vector < ull > H[_];
int n, real, dgr[_], siz[_], mnx, mx[_]; ull h[_], pow[_], v[_];
 
inline void dfs1(int u, int f) {
    siz[u] = dgr[u] != 2, mx[u] = 0;
    for (rg int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].v; if (v == f) continue ;
        dfs1(v, u), siz[u] += siz[v], mx[u] = max(mx[u], siz[v]);
    }
    mx[u] = max(mx[u], real - siz[u]), mnx = min(mnx, mx[u]);
}
 
inline void dfs2(int u, int f) {
    for (rg int i = head[u]; i; i = edge[i].nxt)
        if (edge[i].v != f) dfs2(edge[i].v, u);
    int top = 0;
    for (rg int i = head[u]; i; i = edge[i].nxt)
        if (edge[i].v != f) v[top++] = h[edge[i].v];
    sort(v, v + top), h[u] = top - 1;
    for (rg int i = 0; i < top; ++i) h[u] += pow[i] * v[i];
}
 
vector < int > tmp;
inline bool cmp(const int& i, const int& j) { return s[i] < s[j]; }
 
inline bool check(int x) {
    for (rg int i = 0; i < tmp.size(); ++i) {
    int y = tmp[i];
    for (rg int j = 0; j < H[x].size(); ++j)
        for (rg int k = 0; k < H[y].size(); ++k)
        if (H[x][j] == H[y][k]) return 0;
    }
    return 1;
}
 
int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(m), pow[0] = 1;
    for (rg int i = 1; i < _; ++i) pow[i] = pow[i - 1] * base;    
    for (rg int o = 1; o <= m; ++o) {
        read(n);
        memset(head + 1, tot = 0, sizeof (int) * n);
        memset(dgr + 1, 0, sizeof (int) * n);
        for (rg int u, v, i = 1; i < n; ++i)
            read(u), read(v), Add_edge(u, v), Add_edge(v, u), ++dgr[u], ++dgr[v];
        real = 0;
        for (rg int i = 1; i <= n; ++i) real += dgr[i] != 2;
        mnx = 2e9, dfs1(1, 0);
        s[o] = real, H[o].clear();
        for (rg int i = 1; i <= n; ++i)
            if (mx[i] == mnx) dfs2(i, 0), H[o].push_back(h[i]);
    }
    for (rg int i = 1; i <= m; ++i) if (check(i)) tmp.push_back(i);
    sort(tmp.begin(), tmp.end(), cmp);
    printf("%u
", tmp.size());
    for (rg int i = 0; i < tmp.size(); ++i) printf("%d ", s[tmp[i]]);
    return 0;
}