思路
- 1.先对树从根结点进行一遍dfs求出每个点的深度
- 2.用大根堆把不满足要求的点存起来
- 3.按深度从大到小依次把该点的父亲结点与根结点相连
- 4.更新父亲结点和它的子节点的状态
代码
// Problem: Tree with Small Distances
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/problem/113026
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef long double ld;
const int N=200010;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);
vector<int> g[N];
int dep[N],fa[N],ans;
priority_queue<pii> q;
bool vis[N];
void dfs(int u,int father){
for(int v:g[u]){
if(father==v) continue;
dep[v]=dep[u]+1;
fa[v]=u;
dfs(v,u);
}
}
void solve(){
int n;cin>>n;
_for(i,n-1){
int x,y;cin>>x>>y;
g[x].pb(y);
g[y].pb(x);
}
dfs(1,0);
rep(i,1,n){
if(dep[i]>2) q.push({dep[i],i});
}
while(q.size()){
auto t=q.top();q.pop();
int v=t.Y;
if(vis[v]) continue;
int u=fa[v];
vis[u]=1;
for(int v:g[u]) vis[v]=1;
ans++;
}
cout<<ans;
}
int main(){
ios::sync_with_stdio(0);cin.tie(0);
// int t;cin>>t;while(t--)
solve();
return 0;
}