Codeforces Round #268 (Div. 2) D Two Sets

Two Sets

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little X has n distinct integers: p1, p2, ..., pn. 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, ..., pn (1 ≤ pi ≤ 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, ..., bn (bi equals either0, or 1), describing the division. If bi equals to 0, then pi belongs to set A, otherwise it belongs to set B.

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

Sample test(s)

input

4 5 92 3 4 5

output

YES0 0 1 1

input

3 3 41 2 4

output

NO

Note

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

#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <map>#include <set>#include <queue>#include <stack>#include <bitset>using namespace std;#define PB push_back#define MP make_pair#define REP(i,n) for(int i=0;i<(n);++i)#define FOR(i,l,h) for(int i=(l);i<=(h);++i)#define DWN(i,h,l) for(int i=(h);i>=(l);--i)#define CLR(vis,pos) memset(vis,pos,sizeof(vis))#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LINF 1000000000000000000LL#define eps 1e-8typedef long long ll;int n,a,b;int id;set<int> p;map<int,int> oo;int f[123456];int c[123456];int find_(int x){    if(f[x]==x) return x;    else return f[x]=find_(f[x]);}void union_(int x,int y){    int xx=find_(x);    int yy=find_(y);    if(xx==yy) return;    f[yy]=xx;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    cin>>n>>a>>b;    id=0;    FOR(i,1,n){        scanf("%d",&c[i]);        p.insert(c[i]);        oo[c[i]]=++id;        f[i]=i;    }    f[n+1]=n+1;    f[n+2]=n+2;    FOR(i,1,n){        if(p.count(a-c[i])){            union_(oo[c[i]],oo[a-c[i]]);        }        else union_(oo[c[i]],n+1);        if(p.count(b-c[i])){            union_(oo[c[i]],oo[b-c[i]]);        }        else union_(oo[c[i]],n+2);    }    if(find_(n+1)==find_(n+2)) printf("NO\n");    else{        printf("YES\n");        FOR(i,1,n){            printf("%d ",find_(i)==find_(n+1));        }        puts("");    }    return 0;}