[BJOI2015] 树的同构

Description

Portal

Solution

这题主要就是判断两个树是不是同构。

我们采用Hash来做。

要注意在Hash一个节点的时候要把他的儿子的数量和子树大小hash进去。
因为这是判断两颗树是否同构的重要标准。

此外，我们对于任意的儿子之间最好要有区分。这里采用排序后加权。

Code

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
typedef long long LL;
typedef long double LD;
int read() {
    char ch = getchar();
    int x = 0, flag = 1;
    for (;!isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + 48);
}

const int Maxn = 109, Maxm = 109, Mod = 998244353;
const int LST[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331 ,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503, 509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429, 1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099, 2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741,2749,2753,2767,2777,2789,2791,2797,2801, 2803,2819,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999,3001,3011,3019,3023,3037,3041,3049,3061,3067,3079,3083,3089,3109,3119,3121,3137,3163,3167,3169,3181,3187,3191,3203,3209,3217,3221,3229,3251,3253,3257,3259,3271,3299,3301,3307};
struct Tree {
    struct edge {
        int to, nxt;
    }g[Maxn << 1];
    int head[Maxn], e, size, dep[Maxn], son[Maxn], sz[Maxn];
    LL HashVal[Maxn], zqc[Maxn];
    int randList[Maxn];

    void add(int u, int v) { g[++e] = (edge){v, head[u]}, head[u] = e; }
    
    LL dfsHash(int u, int pa) {
        LL res = 0;
        dep[u] = dep[pa] + 1, son[u] = 0, sz[u] = 1;
        vector <LL> seq; seq.push_back(1);
        for (int i = head[u]; ~i; i = g[i].nxt) 
            if (g[i].to != pa) {
				seq.push_back(dfsHash(g[i].to, u)); 
				++son[u]; sz[u] += sz[g[i].to];
			}
        sort(seq.begin(), seq.end());
        rep (i, 0, seq.size() - 1) (res += seq[i] * randList[i] % Mod) %= Mod;
        return zqc[u] = (res + sz[u] + son[u] + dep[u]) % Mod;
    }

    void init() {
        clar(head, -1), e = 0;
        rep (i, 1, size = read()) {
            int u = read();
            if (u) add(u, i), add(i, u);
        }

        srand(time(NULL));
        rep (i, 1, size) randList[i] = LST[rand() % 100];

        rep (i, 1, size) {
            clar(zqc, 0); clar(dep, 0); clar(son, 0);
            clar(sz, 0); 
            HashVal[i] = dfsHash(i, 0);
        }
    }

    int operator == (const Tree b) const {
        if (size != b.size) return false;
        rep (i, 1, size) 
            rep (j, 1, size)
                if (HashVal[i] == b.HashVal[j]) return 1;
        return 0;
    }
}a[Maxn];

int main() {
	freopen("HashTree.in", "r", stdin);
	freopen("HashTree.out", "w", stdout);

    int m = read();
    rep (i, 1, m) a[i].init();
    rep (i, 1, m) {
        int flag = 0;
        rep (j, 1, i - 1) 
            if (a[i] == a[j]) {
                printf("%d
", j), flag = 1;
                break;
            }
        if (!flag) printf("%d
", i);
    }

#ifdef Qrsikno
    debug("
Running time: %.3lf(s)
", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
    return 0;
}