Codeforces 982C(dfs+思维)

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 $\mathrm{nn vertices.}$

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 $\mathrm{nn (1\le n\le 1051\le n\le 105)\; denoting\; the\; size\; of\; the\; tree.}$

The next $\mathrm{n-1n-1 lines\; contain\; two\; integers uu, vv (1\le u,v\le n1\le u,v\le n)\; each,\; describing\; the\; vertices\; connected\; by\; the ii-th\; edge.}$

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

Output

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

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 $\mathrm{11 and 44.\; The\; graph\; after\; that\; will\; have\; two\; connected\; components\; with\; two\; vertices\; in\; each.}$

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 $-1$

 

二、大致题意

    给出n个点，有n-1条边让他们相连。

    询问最多删除多少边，让他们成为偶数的块

三、思路

    DFS，只要能找到一个偶数的块，就使答案++，因为是偶数块的话，就可以在这里断开一次

(其实感觉一点都不裸，太菜，没想到dfs）

#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#include<vector>
#define ll unsigned long long
#define inf 0x3f3f3f3f
using namespace std;
bool used[100005];
int n;
vector<int>v[100005];
int ans=0;
int dfs(int fa)
{
    used[fa]=1;
    int s=0;
    for(int i=0;i<v[fa].size ();i++)
    {
        if(used[v[fa][i]]) continue;
        s=s+dfs(v[fa][i]);
    }
    if((s+1)%2==0) ans++;//发现是偶数块，就可以减一刀
    return s+1;
}
int main()
{
    int n;
    cin>>n;
    ans=0;
    memset(used,0,sizeof(used));
    for(int i=1;i<=n-1;i++)
    {
        int x,y;
        cin>>x>>y;
        v[x].push_back (y);
        v[y].push_back (x);
    }
    if(n&1) cout<<"-1";
    else
    {
        used[1]=1;
        int s=dfs(1);
        cout<<ans-1;//原本自己就是偶数，所以要减1
    }
    return 0;
}