Vijos1683 有根树的同构问题

• 题目大意: 给出一堆树,求同构(拓扑结构相同)树的集合
• 思路: 一开始写了个前序求置换序列,然后对比后序是否相等,但wa了,还需要对子树进行排序输出其dfs序,但是直接输出按节点多少排序的序列太复杂,于是将一个节点的dfs抽象成\(()\),于是对树\(1 -> 2 , 1 -> 3\)输出的dfs序为\((()())\)

#include<bits/stdc++.h>
#define ll long long 
#define FOR(i,n) for(int i =1; i <= n;++i ) 
#define FOR0(i,n) for(int i =0; i < n;++i )  
#define inf 0x3f3f3f3f
using namespace std; 

int k,n;
int vis[110];
struct node{
	vector<int> son;
	int idx;
	node(){
		idx = 0;
	} 
};
struct Tree{
	int root = 0;
	node nodes[110];
	string pre;
	int in[110];
	int cur = 1;
	int idx;
	string getPre(int root){
//		if(!root)	return ; 
//		getPre(nodes[root].l);
//		getPre(nodes[root].r);
		string z[110];
		int idx = 0;
		for(auto i:nodes[root].son){
			z[++idx] = getPre(i);
		}
		sort(z+1,z+1+idx);
		for(int i=2;i<=idx;++i){
			z[1] = z[1]+z[i];
		}
		return '('+z[1]+')';
	}
	
	bool operator == (Tree const tr)const{
		int len = pre.size();
		FOR0(i,len){
			if(pre[i] != tr.pre[i])	return false;
		}
		return true;
	}
};
vector<Tree> ve;
int main(){
	cin >> k >> n;
	FOR(i,k){
//		cout << i << endl;
		Tree cTree;
		int fr,to;
		FOR(j,n-1){
			cin >> fr >> to;
//			if(cTree.nodes[fr].l == 0){
//				cTree.nodes[fr].l = to;
//			}else{
//				int bro = cTree.nodes[fr].l;
//				while(cTree.nodes[bro].r){
//					bro = cTree.nodes[bro].r;
//				}
//				cTree.nodes[bro].r = to;
//			}
			cTree.nodes[fr].son.push_back(to);
			cTree.nodes[to].idx = 1;
		}
		FOR(j,n){
			if(cTree.nodes[j].idx==0){
				cTree.root =j;
				break;
			}
		}
		cTree.cur = 1;
		cTree.pre = cTree.getPre(cTree.root);
//		cTree.cur = 1;
//		cTree.getIn(cTree.root);
		cTree.idx = i;
		ve.push_back(cTree);
	}
	for(int i=0;i<k;++i){
		if(vis[i])	continue;
		vis[i] = 1;
		cout << ve[i].idx;
		for(int j=i+1;j<k;++j){ 
			if(i==j || vis[j])	continue;
			if(ve[i]==ve[j]){
				cout << '=' << ve[j].idx;
				vis[j] = 1;
			}
		}
		cout << endl;
	}
	for(int i=0;i<k;++i){
		if(!vis[i])	cout << endl << ve[i].idx ;
	}
	return 0;
}

思路来源,
思路来源2