3 solutions
-
0
#include<bits/stdc++.h> using namespace std; const int N=2e5+10; vector<int> g[N]; //g[i]里面储存的是i可以走到的所有点 void dfs(int u,int fa) { cout<<u<<" "; for(auto j:g[u]) { if(j==fa) continue; dfs(j,u); cout<<u<<" "; } } int main() { int n; cin>>n; for(int i=1;i<n;i++) { int a,b; cin>>a>>b; g[a].push_back(b); g[b].push_back(a); } for(int i=1;i<=n;i++) { sort(g[i].begin(),g[i].end()); } dfs(1,-1);//当前点和当前点的父节点 return 0; }
Information
- ID
- 2772
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 3
- Tags
- # Submissions
- 14
- Accepted
- 9
- Uploaded By