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;
}