题目大意：给出n个数，要求将n个数分配到两个集合中，集合0中的元素x，要求A-x也再0中，同理1集合。

写了几个版本号，一直WA在第8组数据...最后參考下ans，写了并查集过了

学到：1、注意离散的逻辑思维，官方答案的 从条件推逆否命题

2、并查集做法：fa[find(i)]=mp[a-p[i]]
? find(a-p[i]) :
find(n+2);

3、离散化然后hash的方法，用map时间还是承受得住的，写起来也简单

//#pragma comment(linker, "/STACK:102400000,102400000")

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <iostream>

#include <iomanip>

#include <cmath>

#include <map>

#include <set>

#include <queue>

using namespace std;

#define ls(rt) rt*2

#define rs(rt) rt*2+1

#define ll long long

#define ull unsigned long long

#define rep(i,s,e) for(int i=s;i<e;i++)

#define repe(i,s,e) for(int i=s;i<=e;i++)

#define CL(a,b) memset(a,b,sizeof(a))

#define IN(s) freopen(s,"r",stdin)

#define OUT(s) freopen(s,"w",stdout)

const ll ll_INF = ((ull)(-1))>>1;

const double EPS = 1e-8;

const double pi = acos(-1.0);

const int INF = 100000000;

const int MAXN = 1e5+100;

int fa[MAXN],p[MAXN],vis[MAXN];

int n,a,b;

map <int , int> mp;

int Fi(int x)

{

    if(x!=fa[x])fa[x]=Fi(fa[x]);

    return fa[x];

}

void un(int x, int y)

{

    x=Fi(x);

    y=Fi(y);

    fa[x]=y;

}

int flag;

void init()

{

    for(int i=0;i<=n+3;i++)

        fa[i]=i;

    mp.clear();

    CL(vis,0xff);

    flag=1;

}

/*void solve()

{

    for(int i=1;i<=n;i++)

    {

        if(mp[a-p[i]])

        {

            if(Fi(fa[mp[a-p[i]]])==Fi(n+2) || mp[b-p[i]])

            {

                flag=0;

                return;

            }

            else

            {

                un(mp[a-p[i]],n+1);

            }

        }

        else

        {

            if(mp[b-p[i]])

            {

                if(Fi(fa[mp[b-p[i]]]) == Fi(n+1) || mp[a-p[i]])

                {

                    flag=0;

                   return;

                }

                else

                {

                    un(mp[b-p[i]],n+2);

                }

            }

            else

            {

                flag=0;

                return;

            }

        }

    }

}*/

void solve()

{

    for(int i=1;i<=n;i++)

    {

        fa[Fi(i)] = mp[a-p[i]] ? Fi(mp[a-p[i]]) : Fi(n+2);

        fa[Fi(i)] = mp[b-p[i]] ? Fi(mp[b-p[i]]) : Fi(n+1);

    }

}

int main()

{

    //IN("D.txt");

    while(~scanf("%d%d%d",&n,&a,&b))

    {

        init();

        for(int i=1;i<=n;i++)

        {

            scanf("%d",&p[i]);

            mp[p[i]]=i;

        }

        solve();

        if(flag && Fi(n+1)!=Fi(n+2))

        {

            printf("YES\n");

            int ans[MAXN],cnt=1;

            for(int i=1;i<=n;i++)

            {

                if(Fi(i) == Fi(n+1))

                    ans[i]=0;

                else

                    ans[i]=1;

            }

            printf("%d",ans[1]);

            for(int i=2;i<=n;i++)

                printf(" %d",ans[i]);

            putchar('\n');

        }

        else

            puts("NO");

    }

    return 0;

}