Cut 'em all! CodeForces - 982C（dfs）

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 $-1$

题意：有一颗树，现在，你需要切掉一些边，使这颗树分成若干个节点为偶数的部分，问，最多能切掉几条边。不能切,输出-1。

思路：首先，只有节点数为偶数的才能分成若干个偶数的部分，所以节点数为奇数的就直接输出-1吧。如果是偶数的话，从任意一点进行dfs，dfs返回的是子树节点的个数，如果是偶数的话，那么，父节点与子树的这条边一定是可以切掉的，所以，只要记录子树节点为偶数的个数就可以了。

#include "iostream"
#include "vector"
#include "cstdio"
using namespace std;
const int Max=1e5+10;
vector<int > G[Max];
bool flag[Max];//避免重复计算
int ans;
int dfs(int x)
{
    int res=0;
    for(int i=0;i<G[x].size();i++){
        int v=G[x][i];
        if(!flag[v]){
            flag[v]=1;
            int tmp=dfs(v);
            if(tmp%2==0) ans++;//如果子树节点的个数是偶数，那么，这条边就可以切掉
            res+=tmp;
        }
    }
    return res+1;//子树节点加一，就是这颗树的节点数
}
int main()
{
    int n,u,v;
    ans=0;
    scanf("%d",&n);
    for(int i=0;i<n-1;i++){
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    if(n%2!=0)//如果节点数为奇数，那么一定不可以切边
        printf("-1\n");
    else{
        flag[1]=1;
        dfs(1);
        printf("%d\n",ans);
    }
    return 0;
}