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，一直WA在第8组数据...最后参考下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;
}

Codeforces 468B Two Sets 并查集