HDU 4013 Distinct Subtrees(树的最小表示)

Distinct Subtrees

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 391    Accepted Submission(s): 190

Problem Description

Given an unrooted tree with n nodes, the subtree is defined as a connected component of the given tree. Two subtrees are considered the same if there exists a map from nodes of one tree to another so that the edges of both trees are corresponding the same pair of nodes after mapping.
  Your task is to find out how many distinct subtrees for a given unrooted tree.

 

Input

The input consists of multiple test cases. The first line of input contains an integer denoting the number of test cases.
  For each test case, the first line contains one integer n denoting the number of nodes of the given tree. (1 <= n <= 15)
  Then n-1 lines follow, each line contains two integers denoting an edge of the tree (x, y).

 

Output

For each test case, output the number of distinct subtrees for the given tree.

 

Sample Input

2

3

1 2

1 3

9

9 4

4 3

1 3

7 4

1 6

5 7

2 4

6 8

 

Sample Output

Case #1: 3

Case #2: 21

 

 

问一棵树有多少棵结构不同的子树。

做法是将它dfs形式最小表示后，插入trie中

 

#include<bits/stdc++.h>
using namespace std ;
const int N = 22 ;
int n , g[N][N] , a[N] , b[N] , st , ans ;

struct trie {
    int date;
    struct trie* son[2];
}*root;

void init() {
    ans = 0 ;
    memset( g , 0 , sizeof g ) ;
    root = new trie ;
    root -> date = 0 ;
    root -> son[0] = NULL;
    root -> son[1] = NULL;
}
int insert( string s , struct trie *p ) {
    struct trie *rot = p ;
    for( int i = 0 ; i < s.size() ; ++i ) {
        if( rot -> son[ s[i] - '0'] == NULL ) {
            trie *t = new trie ;
            t -> date = 0 ;
            t -> son[0] = NULL ;
            t -> son[1] = NULL ;
            rot -> son[ s[i] - '0' ] = t ;
        }
        rot = rot -> son[ s[i] - '0' ] ;
    }
    if( rot -> date == 1 ) return 0 ;
    rot -> date = 1 ;
    return 1 ;
}

string dfs1( int u , int p ) {
    string vs = "0";
    vector<string>q;
    for( int i = 0 ; i < n ; ++i )
        if( (st&(1<<i)) && g[u][i] && i != p )
            q.push_back(dfs1(i,u));
    sort( q.begin() , q.end() );
    for( int i = 0 ; i < q.size() ; ++i ) vs += q[i] ;
    vs += "1";
    return vs ;
}

int solve() {
    int f = 0 , t ;
    string s ;
    for( int i = 0 ; i < n ; ++i ) if( st&(1<<i) ){
        s = dfs1( i , -1 );
        t = insert( s , root );
        if( (!t)&& (!f) ) return 0 ;
        f = 1 ;
    }
    return 1 ;
}

int main() {
    string s ;
    ios::sync_with_stdio(0);
    int _ , cas = 1 ; cin >> _ ;
    while( _-- ) {
        cin >> n ;
        init() ;
        for( int i = 1; i < n ; ++i ) {
            int u , v ; cin >> u >> v ;
            u-- , v-- ;
            g[u][v] = g[v][u] = 1 ;
        }
        for( int i = 1 ; i < (1<<n) ; ++i ) {
            st = i ;
            ans += solve();
        }
        cout << "Case #"<< cas++ << ": " << ans << endl ;
    }
    return 0 ;
}