bzoj4337: BJOI2015 树的同构 树哈希判同构

题目链接

bzoj4337: BJOI2015 树的同构

题解

树哈希的一种方法
对于每各节点的哈希值为hash[x] = hash[sonk[x]] * p[k];
p为素数表

代码

#include<cstdio> 
#include<cstring> 
#include<algorithm> 
 
inline int read() { 
    int x = 0,f = 1; 
    char c  = getchar() ; 
    while(c < '0' || c > '9')c = getchar(); 
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
    return x * f; 
} 
const int maxn = 3007; 
unsigned int hash[maxn][maxn]; 
unsigned int p[]={0,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317};
unsigned int f[maxn]; 
int m; 
struct node { 
    int v,next; 
}  edge[maxn << 1]; 
int num = 0; 
int head[maxn];  
inline void add_edge(int u,int v) {edge[++ num].v = v; edge[num].next = head[u];head[u] = num;} 
void dfs(int x,int fa) { 
    int num = 0; 
    unsigned int son[107]; 
    son[++ num] = 1; 
    for(int i = head[x];i;i = edge[i].next) { 
        int v = edge[i].v; 
        if(v == fa) continue; 
        dfs(v,x); 
        son[++ num] = f[v]; 
    } 
    std::sort(son + 1, son + num + 1); 
    f[x] = 0; 
    for(int i = 1;i <= num;++ i) f[x] += p[i] * son[i]; 
} 
int main() { 
    m = read(); 
    int n; 
    for(int i = 1;i <= m;++ i) { 
        memset(head,0,sizeof head); num = 0; 
        n = read(); 
        for(int x,j = 1;j <= n;++ j) { 
            x = read(); if(x) { add_edge(x,j),add_edge(j,x); } 
        } 
        for(int j = 1;j <= n;++ j) { dfs(j,0),hash[i][j] = f[j];} 
        std::sort(hash[i] + 1,hash[i] + n + 1); 
        for(int j = 1;j <= i;++ j) { 
            int k; 
            for(k = 1;k <= n;++ k) 
                if(hash[j][k] != hash[i][k]) break; 
            if(k > n) {printf("%d\n",j);break; } 
        } 
    } 
    return 0; 
}