Tree with Small Distances

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;
}