BZOJ4337: BJOI2015 树的同构(hash 树同构)

题意

题目链接

Sol

树的同构问题，直接拿hash判一下，具体流程大概是这样的：

首先转化为有根树，预处理出第(i)棵树以(j)为根时的hash值。

那么两个树同构当且仅当把两棵树的hash数组排完序后完全一致(感性理解一下)

/*

*/
#include<bits/stdc++.h> 
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long 
#define ull unsigned long long 
using namespace std;
const int MAXN = 51, mod = 1e9 + 7;
const ull base = 997;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, ha[MAXN][MAXN], fa[MAXN], ans[MAXN], top, num[MAXN];
ull st[MAXN], f[MAXN];
vector<int> v[MAXN];
ull dfs(int x, int fa) {
    vector<int> st;   
    f[x] = 1;
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(to == fa) continue;
        st.push_back(dfs(to, x));
    }
    sort(st.begin(), st.end());
    for(int i = 0; i < st.size(); i++) f[x] = base * f[x] + st[i];
    return f[x];
}
bool check(int a, int b) {
    if(num[a] != num[b]) return 0;
    for(int i = 1; i <= num[a]; i++)
        if(ha[a][i] != ha[b][i]) return 0;
    return 1;
}
signed main() {
    N = read();
    for(int i = 1; i <= N; i++) {
        num[i] = read();
        for(int j = 1; j <= num[i]; j++) v[j].clear();
        for(int j = 1; j <= num[i]; j++) {
            fa[j] = read();
            if(fa[j]) v[fa[j]].push_back(j), v[j].push_back(fa[j]);
        }
        for(int j = 1; j <= num[i]; j++)
            ha[i][j] = dfs(j, 0);
        sort(ha[i] + 1, ha[i] + num[i] + 1);
    }
    for(int i = 1; i <= N; i++) {
        ans[i] = i;
        for(int j = 1; j <= i - 1; j++) 
            if(check(j, i)) {ans[i] = j; break;}
    }
    for(int i = 1; i <= N; i++) printf("%d
", ans[i]);
    return 0;
}
/*
4 
4 2 0 2 3 
4 0 1 1 2
4 0 1 1 1 
4 0 1 2 3 
*/