- 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