C. Bakry and Partitioning(Codeforces Round #746 (Div. 2))

链接:C. Bakry and Partitioning.
思路:a^a=0,dfs找出子树异或和等于sum的个数。

int t,n,k;
int a[100010],b[100010],f[100010];
vectorc[100010];
int ans,sum;
void dfs(int u)
{
    f[u]=1;
    for(int i=0;i>t;
    while(t--)
    {
        ans=0;sum=0;
        cin>>n>>k;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i],b[i]=a[i],c[i].clear();
            sum=sum^a[i];f[i]=0;
        }
        for(int i=1;i>u>>v;
            c[u].push_back(v);
            c[v].push_back(u);
        }
        if(sum==0){cout<<"YESn";continue;}
        if(k==2){cout<<"NOn";continue;}
        dfs(1);
        if(ans==1)cout<<"NOn";
        else cout<<"YESn";
    }
    return 0;
}