Los homogeneous solution to a problem tree valley P5043

Face questions

 

The difficulty of this question is actually not as difficult questions of purple. Mainly in the details of processing more complicated when a hash;

The first is the tree hash template:

long long treehash(int u,int fa)
{
    long long q[1001];
    long long num=0;
    long long ans=1;
    for(int i=head[u];i;i=star[i].nxt){
        int v=star[i].to;
        if(v==fa) continue;
        q[++num]=treehash(v,u);
    }
    sort(q+1,q+num+1);
    for(int i=1;i<=num;i++){
        ans=ans*2333+q[i];
    }
    return ans*2333+1;
}

For unrooted trees in little time data may be sequentially enumerated root hash value for each point of time when the root, then the tree becomes a hash value of the hash value of a set of numbers;

N ^ 2 and then compare each linear tree can be a hash value;

 

#include <bits/stdc++.h>
using namespace std;
struct littlestar{
    int to;
    int nxt;
}star[20010];
long long head[20010],cnt;
void add(int u,int v)
{
    star[++cnt].to=v;
    star[cnt].nxt=head[u];
    head[u]=cnt;
}
long long myhash[20010];
long long treehash(int u,int fa)
{
    long long q[1001];
    long long num=0;
    long long ans=1;
    for(int i=head[u];i;i=star[i].nxt){
        int v=star[i].to;
        if(v==fa) continue;
        q[++num]=treehash(v,u);
    }
    sort(q+1,q+num+1);
    for(int i=1;i<=num;i++){
        ans=ans*2333+q[i];
    }
    return ans*2333+1;
}
long long lala[20010];
long long ans[20010];
int main ()
{
    int t;
    cin>>t;
    for(int i=1;i<=t;i++){
        memset(head,0,sizeof(head));
        memset(lala,0,sizeof(lala));
        cnt=0;
        int n;
        cin>>n;
        int root=1;
        for(int j=1;j<=n;j++){
            int x;
            scanf("%d",&x);
            if(x==0){
                root=j;
                continue;
            } 
            add(x,j);
            add(j,x);        
        }
        for(intj = 1 ; j <= n; j ++) myhash [j] = treehash (j, 0 ); 
        sort (myhash + 1 , myhash + 1 + n);
        for ( int j = 1 ; j <= n; j ++) lala [j] = lala [j- 1 ] * myhash [j] + 233 ; 
        years [i] = lala [n];
        for ( int j = 1 ; j <= i; j ++ ) {
             if (years [i] == years [j]) { 
                cout << d << endl;
                break ; 
            } 
        }
    }
    return 0;
}