Codeforces Round #484 (Div. 2) C. Cut 'em all!

Example 1
input
4
2 4
4 1
3 1
output
1

Example 2
input
3
1 2
1 3
output
-1

Example 3
input
10
7 1
8 4
8 10
4 7
6 5
9 3
3 5
2 10
2 5
output
4


Example 4
input
2
1 2
output
0

　　

题目大意：

n个顶点，n-1一条边，将其中的边删除，然遗留下的边都包含偶数个点，求最多可以删多少条边

分析：

深搜，只有某一个点的儿子为奇数，ans++

code:

#define debug
#include<bits/stdc++.h>
#define pb push_back
#define dbg(x) cout<<#x<<" = "<<(x)<<endl;
#define lson l,m,rt<<1
#define cmm(x) cout<<"("<<(x)<<")";
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>PLL;
typedef pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const ll inf=0x7fffffff;
const double eps=1e-8;
const int maxn =1e6+10;
const int N = 510;
const ll mod=1e9+7;
const ll MOD=1e9;
//------
//define
vector<int>G[maxn];
int subtree[maxn];
int vis[maxn];
int ans;
//dfs
void dfs(int u){
	vis[u]=1;
	for(int i=0;i<G[u].size();i++){
		if(vis[G[u][i]])continue;
		dfs(G[u][i]);
		subtree[u]+=subtree[G[u][i]];
		if(subtree[G[u][i]]%2==0){
			ans++;
		}
	}
	subtree[u]+=1;
}
//solve
void solve() {
	int n;
	while(cin>>n){
		for(int i=0;i<n+2;i++){
			G[i].clear();
		}
		for(int i=0;i<n-1;i++){
			int u,v;
			cin>>u>>v;
			G[v].push_back(u);
			G[u].push_back(v);
		}
		if(n%2){
			cout<<-1<<endl;
			continue;
		}
		ans=0;
		dfs(1);
		cout<<ans<<endl;
		memset(vis,0,sizeof(vis));
		memset(subtree,0,sizeof(subtree));
	}
}

int main() {
	ios_base::sync_with_stdio(false);
#ifdef debug
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	cin.tie(0);
	cout.tie(0);
	solve();
	/*
		#ifdef debug
			fclose(stdin);
			fclose(stdout);
			system("out.txt");
		#endif
	*/
	return 0;
}