AtCoder Grand Contest 014 B

B - Unplanned Queries

Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

Takahashi is not good at problems about trees in programming contests, and Aoki is helping him practice.

First, Takahashi created a tree with N vertices numbered 1 through N, and wrote 0 at each edge.

Then, Aoki gave him M queries. The i-th of them is as follows:

Increment the number written at each edge along the path connecting vertices ai and bi, by one.

After Takahashi executed all of the queries, he told Aoki that, for every edge, the written number became an even number. However, Aoki forgot to confirm that the graph Takahashi created was actually a tree, and it is possible that Takahashi made a mistake in creating a tree or executing queries.

Determine whether there exists a tree that has the property mentioned by Takahashi.

Constraints

2≤N≤105
1≤M≤105
1≤ai,bi≤N
ai≠bi

Input

Input is given from Standard Input in the following format:

N M
a1 b1
:
aM bM

Output

Print YES if there exists a tree that has the property mentioned by Takahashi; print NO otherwise.

Sample Input 1

Copy

Sample Output 1

Copy

YES

For example, Takahashi's graph has the property mentioned by him if it has the following edges: 1−2, 1−3 and 1−4. In this case, the number written at every edge will become 2.

Sample Input 2

Copy

Sample Output 2

Copy

NO

如果成立那么树的叶子节点的的度必定为偶数，那么删掉所有叶子节点和连接叶子节点的边，又是一个树，同理可得到树的所有节点的必定被提到偶数次

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))
using namespace std;
const int inf=1e9+10;
const ll llinf=1e16+10;
const int maxn=1e5+10;
const int maxm=1e2+10;
const int mod=1e9+7;
int n,m;
int c[maxn];
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            c[a]++;
            c[b]++;
        }
        int flag=1;
        for(int i=1;i<=n;i++)
        {
            if(c[i]%2==1)
                flag=0;
        }
        if(flag) printf("YES
");
        else printf("NO
");
    }
    return 0;
}