CF1029E Tree with Small Distances

思路:我怎么感觉好像在蓝书见过你.
$dp(u,0):根直接与u连接的花费$
$dp(u,1):u与根距离为2而且父亲与根相连$
$dp(u,2):u与根距离为2且儿子中有一个与根相连$
$$dp(u,0)=1+\sum min(dp(v,0),dp(v,1),dp(v,2))$$
$$dp(u,1)=\sum min(dp(v,0),dp(v,2))$$
$dp(u,2)$需要至少有一个儿子节点是$dp(v,0)$状态,其余儿子可以是$dp(v,0)或dp(v,2)$

#include
using namespace std;
const int maxn = 1e6+5;
const int INF = 2e5+5;
typedef long long ll;
typedef pair<int,int> pii;
#define all(a) (a).begin(), (a).end()
#define pb(a) push_back(a)
vector<int> G[maxn];
int dp[maxn][3];
int dfs(int u,int fa,int st){
	int &ans = dp[u][st];
	if(ans!=-1) return ans;
	ans = 0;
	if(st==0) ans++;
	int child = 0;
	vector<pii> res;
	for(auto v : G[u]){
		if(v==fa) continue;
		child++;
		if(u==1){
			ans += dfs(v,u,0) - 1;
		}
		else{
			if(st==0) ans+= min(dfs(v,u,0),min(dfs(v,u,1),dfs(v,u,2)));
			else if(st==1) ans+=min(dfs(v,u,0),dfs(v,u,2));
			else if(st==2) res.push_back({dfs(v,u,0),dfs(v,u,2)});
		}
	}
	if(child==0&&st==2){
		ans = INF;
		return ans;
	}
	if(st==2){
		ans = INF;int r = 0;
		for(auto [a,b] : res) r += min(a,b);
		for(auto [a,b] : res) ans = min(ans,r - min(a,b) + a);
	}
	return ans;
}
int main(){
    ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	memset(dp,-1,sizeof(dp));
	int n;cin>>n;
	for(int i=1;i<=n-1;i++){
		int u,v;cin>>u>>v;
		G[u].pb(v);G[v].pb(u);
	}
	cout<<dfs(1,0,1)<<"\n";
}
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