树hash
判断树的同构，采用树hash的方式。
树hash定义在有根树上。判断无根树同构的时候，可以比较重心为根的hash值或者比较每个点为根的hash值。
h[x]表示x为根的子树的hash，g[x]表示x为根时全树的hash。
我采用的方法是
h[x] = 1 + ∑h[y] * p[siz[y]]
于是g[x] = g[fa] - h[x] * p[siz[x]] + h[x]
例题1: BJOI2015 树的同构
判断无根树同构，我是比较了每个点为根时的hash值。
1 #include <bits/stdc++.h> 2 3 typedef long long LL; 4 const int N = 60, MO = 998244353; 5 6 struct Edge { 7 int nex, v; 8 }edge[N << 1]; int tp; 9 10 int e[N], n, m, turn, fr[N], p[1000010], top, siz[N], h[N], g[N]; 11 std::vector<int> v[N]; 12 bool vis[1000010]; 13 14 inline void add(int x, int y) { 15 tp++; 16 edge[tp].v = y; 17 edge[tp].nex = e[x]; 18 e[x] = tp; 19 return; 20 } 21 22 inline bool equal(int a, int b) { 23 int len = v[a].size(); 24 if(len != v[b].size()) return false; 25 for(int i = 0; i < len; i++) { 26 if(v[a][i] != v[b][i]) return false; 27 } 28 return true; 29 } 30 31 inline void getp(int n) { 32 for(int i = 2; i <= n; i++) { 33 if(!vis[i]) { 34 p[++top] = i; 35 } 36 for(int j = 1; j <= top && i * p[j] <= n; j++) { 37 vis[i * p[j]] = 1; 38 if(i % p[j] == 0) { 39 break; 40 } 41 } 42 } 43 return; 44 } 45 46 void DFS_1(int x, int f) { 47 siz[x] = 1; 48 h[x] = 1; 49 for(int i = e[x]; i; i = edge[i].nex) { 50 int y = edge[i].v; 51 if(y == f) continue; 52 DFS_1(y, x); 53 h[x] = (h[x] + 1ll * h[y] * p[siz[y]] % MO) % MO; 54 siz[x] += siz[y]; 55 } 56 return; 57 } 58 59 void DFS_2(int x, int f, int V) { 60 g[x] = (h[x] + 1ll * V * p[n - siz[x]] % MO) % MO; 61 v[turn].push_back(g[x]); 62 V = (1ll * V * p[n - siz[x]] % MO + 1) % MO; 63 for(int i = e[x]; i; i = edge[i].nex) { 64 int y = edge[i].v; 65 if(y == f) { 66 continue; 67 } 68 DFS_2(y, x, ((LL)V + h[x] - 1 - 1ll * h[y] * p[siz[y]] % MO + MO) % MO); 69 } 70 return; 71 } 72 73 int main() { 74 getp(1000009); 75 scanf("%d", &m); 76 for(turn = 1; turn <= m; turn++) { 77 scanf("%d", &n); 78 tp = 0; 79 memset(e + 1, 0, n * sizeof(int)); 80 for(int i = 1, x; i <= n; i++) { 81 scanf("%d", &x); 82 if(x) { 83 add(x, i); 84 add(i, x); 85 } 86 } 87 DFS_1(1, 0); 88 DFS_2(1, 0, 0); 89 std::sort(v[turn].begin(), v[turn].end()); 90 /*for(int i = 0; i < n; i++) { 91 printf("%d ", v[turn][i]); 92 } 93 puts("\n");*/ 94 } 95 96 for(int i = 1; i <= m; i++) { 97 fr[i] = i; 98 } 99 for(int i = 2; i <= m; i++) { 100 for(int j = 1; j < i; j++) { 101 if(equal(i, j)) { 102 fr[i] = fr[j]; 103 break; 104 } 105 } 106 } 107 for(int i = 1; i <= m; i++) { 108 printf("%d\n", fr[i]); 109 } 110 return 0; 111 }
例题2: JSOI2016 独特的树叶
对第一棵树的所有点为根的hash值建立set，然后枚举第二棵树，在set中查。
1 #include <bits/stdc++.h> 2 3 const int N = 100010, MO = 998244353; 4 5 struct Edge { 6 int nex, v; 7 }; 8 9 std::set<int> st; 10 int p[2000010], top, in[N], near[N]; 11 bool vis[2000010]; 12 13 inline int qpow(int a, int b) { 14 int ans = 1; 15 while(b) { 16 if(b & 1) ans = 1ll * ans * a % MO; 17 a = 1ll * a * a % MO; 18 b = b >> 1; 19 } 20 return ans; 21 } 22 23 struct Tree { 24 Edge edge[N << 1]; int tp; 25 int e[N], h[N], g[N], siz[N], n; 26 inline void init(int t) { 27 n = t; 28 tp = 0; 29 memset(e + 1, 0, n * sizeof(int)); 30 return; 31 } 32 inline void add(int x, int y) { 33 edge[++tp].v = y; 34 edge[tp].nex = e[x]; 35 e[x] = tp; 36 return; 37 } 38 void DFS_1(int x, int f) { 39 siz[x] = 1; 40 h[x] = 1; 41 for(int i = e[x]; i; i = edge[i].nex) { 42 int y = edge[i].v; 43 if(y == f) { 44 continue; 45 } 46 DFS_1(y, x); 47 siz[x] += siz[y]; 48 h[x] = (h[x] + 1ll * h[y] * p[siz[y]] % MO) % MO; 49 } 50 //printf("x = %d h[x] = %d \n", x, h[x]); 51 return; 52 } 53 void DFS_2(int x, int f, int V) { 54 g[x] = (h[x] + 1ll * V * p[n - siz[x]] % MO) % MO; 55 //printf("x = %d v = %d g = %d \n", x, V, g[x]); 56 for(int i = e[x]; i; i = edge[i].nex) { 57 int y = edge[i].v; 58 if(y == f) { 59 continue; 60 } 61 DFS_2(y, x, (g[x] - 1ll * h[y] * p[siz[y]] % MO + MO) % MO); 62 } 63 return; 64 } 65 }t0, t1; 66 67 inline void getp(int n) { 68 for(int i = 2; i <= n; i++) { 69 if(!vis[i]) { 70 p[++top] = i; 71 } 72 for(int j = 1; j <= top && i * p[j] <= n; j++) { 73 vis[i * p[j]] = 1; 74 if(i % p[j] == 0) { 75 break; 76 } 77 } 78 } 79 return; 80 } 81 82 int main() { 83 84 getp(2000009); 85 86 int n; 87 scanf("%d", &n); 88 t0.init(n); 89 t1.init(n + 1); 90 int x, y; 91 for(int i = 1; i < n; i++) { 92 scanf("%d%d", &x, &y); 93 t0.add(x, y); 94 t0.add(y, x); 95 } 96 for(int i = 1; i <= n; i++) { 97 scanf("%d%d", &x, &y); 98 t1.add(x, y); 99 t1.add(y, x); 100 in[x]++; 101 in[y]++; 102 near[x] = y; 103 near[y] = x; 104 } 105 106 t0.DFS_1(1, 0); 107 t0.DFS_2(1, 0, 0); 108 for(int i = 1; i <= n; i++) { 109 st.insert(t0.g[i]); 110 //printf("%d hash = %d \n", i, t0.g[i]); 111 } 112 113 t1.DFS_1(1, 0); 114 t1.DFS_2(1, 0, 0); 115 for(int i = 1; i <= n + 1; i++) { 116 if(in[i] == 1) { 117 int x = (t1.g[near[i]] - 2 + MO) % MO; 118 if(st.find(x) != st.end()) { 119 printf("%d\n", i); 120 return 0; 121 } 122 } 123 } 124 125 return 0; 126 }
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