Codeforces 982C(dfs)

传送门

题面：

C. Cut 'em all!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You're given a tree with $n$ vertices.

Your task is to determine the maximum possible number of edges that can be removed in such a way that all the remaining connected components will have even size.

Input

The first line contains an integer $n$ ($1\le n\le {10}^5$) denoting the size of the tree.

The next $n-1$ lines contain two integers $u$, $v$ ($1\le u,v\le n$) each, describing the vertices connected by the $i$-th edge.

It's guaranteed that the given edges form a tree.

Output

Output a single integer $k$ — the maximum number of edges that can be removed to leave all connected components with even size, or $-1$if it is impossible to remove edges in order to satisfy this property.

Examples

input

Copy

4
2 4
4 1
3 1

output

Copy

1

input

Copy

3
1 2
1 3

output

Copy

-1

input

Copy

10
7 1
8 4
8 10
4 7
6 5
9 3
3 5
2 10
2 5

output

Copy

4

input

Copy

2
1 2

output

Copy

0

Note

In the first example you can remove the edge between vertices $1$ and $4$. The graph after that will have two connected components with two vertices in each.

In the second example you can't remove edges in such a way that all components have even number of vertices, so the answer is $-1$.

题目描述：

    给你一个有n个结点的树，其中有n-1条边，询问你最多你可以移动多少条边，使得每个强连通分量的个数均为偶数。

题目分析：

    因为我们知道，如果所给的结点的个数是奇数，那么，因为奇数必定是由一个偶数和一个奇数组成，故总会有一个强连通分量的个数为奇数，因此当n为奇数使，答案为0。

    那么当n等于偶数的时，我们只需要进行dfs遍历整张图，统计一下某个结点以及其儿子结点的数量，如果递归到最底时统计出的数量为偶数，则答案+1即可。

代码：

#include <bits/stdc++.h>
#define maxn 200005
using namespace std;
struct edge{
    int to,next;
}q[maxn];
int head[maxn];
int sizz[maxn];
int cnt=0;
int ans=0;
void init(){
    memset(head,-1,sizeof(head));
    cnt=0;
}
void add_edge(int from,int to){
    q[cnt].next=head[from];
    q[cnt].to=to;
    head[from]=cnt++;
}
void dfs(int x,int fa){
    sizz[x]=1;
    for(int i=head[x];i!=-1;i=q[i].next){
        if(q[i].to==fa) continue;
        dfs(q[i].to,x);
        sizz[x]+=sizz[q[i].to];
    }
    if(sizz[x]%2==0) ans++,sizz[x]=0;
}
int main()
{
    int n;
    cin>>n;
    init();
    for(int i=0;i<n-1;i++){
        int a,b;
        cin>>a>>b;
        add_edge(a,b);
        add_edge(b,a);
    }
    if(n&1) {
        puts("-1");
        return 0;
    }
    dfs(1,-1);
    cout<<ans-1<<endl;
}