cf468B Two Sets

Posted 2020-09-28 zhber

Little X has n distinct integers: p₁,?p₂,?...,?p_n. He wants to divide all of them into two sets A and B. The following two conditions must be satisfied:

• If number x belongs to set A, then number a?-?x must also belong to set A.
• If number x belongs to set B, then number b?-?x must also belong to set B.

Help Little X divide the numbers into two sets or determine that it‘s impossible.

Input

The first line contains three space-separated integers n,?a,?b (1?≤?n?≤?10⁵; 1?≤?a,?b?≤?10⁹). The next line contains n space-separated distinct integers p₁,?p₂,?...,?p_n (1?≤?p_i?≤?10⁹).

Output

If there is a way to divide the numbers into two sets, then print "YES" in the first line. Then print n integers: b₁,?b₂,?...,?b_n (b_i equals either 0, or 1), describing the division. If b_i equals to 0, then p_i belongs to set A, otherwise it belongs to set B.

If it‘s impossible, print "NO" (without the quotes).

Examples

Input

4 5 9
2 3 4 5

Output

YES
0 0 1 1

Input

3 3 4
1 2 4

Output

NO

Note

It‘s OK if all the numbers are in the same set, and the other one is empty.

 

如果A中有了一个x，那么A中也要有a-x，说明（逆否命题）如果A中没有a-x，也就没有x。即如果B中有a-x，就有x。因为不在A就在B咯

所以这个A还是B无所谓的，重要的是x和a-x一定在同一集合，x和b-x一定在同一集合

因此裸并查集，只要被并起来的数字有一个不能在A，那么这一群都不能在A

 1 #include<bits/stdc++.h>
 2 #define LL long long
 3 using namespace std;
 4 LL n,a,b;
 5 inline LL read()
 6 {
 7     LL x=0,f=1;char ch=getchar();
 8     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 9     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
10     return x*f;
11 }
12 struct po{LL x;int rnk;}p[100010];
13 bool operator <(po a,po b){return a.x<b.x;}
14 int fa[100010];
15 int mrk[100010];
16 inline int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
17 map<LL,LL>mp;
18 int main()
19 {
20     n=read();a=read();b=read();
21     for (int i=1;i<=n;i++)
22     {
23         p[i].x=read();
24         p[i].rnk=i;
25     }
26     sort(p+1,p+n+1);
27     for(int i=1;i<=n;i++)mp[p[i].x]=p[i].rnk,fa[i]=i,mrk[i]=3;
28     for(int i=1;i<=n;i++)
29     {
30         if (mp[a-p[i].x])
31         {
32             int pos=getfa(mp[a-p[i].x]),pos2=getfa(p[i].rnk);
33             if (pos!=pos2)fa[pos2]=pos;
34         }
35         if (mp[b-p[i].x])
36         {
37             int pos=getfa(mp[b-p[i].x]),pos2=getfa(p[i].rnk);
38             if (pos!=pos2)fa[pos2]=pos;
39         }
40     }
41     for (int i=1;i<=n;i++)
42     {
43         int ff=getfa(p[i].rnk);
44         if (!mp[a-p[i].x])mrk[ff]&=1;
45         if (!mp[b-p[i].x])mrk[ff]&=2;
46     }
47     for (int i=1;i<=n;i++)
48         if (mrk[getfa(i)]==0){puts("NO");return 0;}
49         else if (mrk[getfa(i)]==3)mrk[getfa(i)]=1;
50     puts("YES");
51     for (int i=1;i<=n;i++)printf("%d ",mrk[getfa(i)]==2?0:1);
52     puts("");
53 }

cf468B

 

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