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本文介绍了一种将一组整数划分为两个集合A和B的方法，遵循特定规则：若x属于A，则a-x也必须属于A；若x属于B，则b-x也必须属于B。通过DFS算法解决该问题，并提供完整的C++实现代码。

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题意:

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 either 0, 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).

题解:

因为一个数要么A要么属于B..那么将它们都先定为A..再把不符合的调整到B..由于改变了一个位置从A到B..会让若干的数也会变..用一个dfs来处理就好...

Program:

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h> 
#include<string>
#include<queue>
#include<map>
#include<algorithm>
#define MAXN 100005
#define ll long long
#define oo 1e+10
#define eps 1e-10
using namespace std;  
struct node
{
      int w,id;
      bool operator<(node a) const
      {
              return w<a.w;
      }
}P[MAXN];
int ans[MAXN],n,a,b; 
int find(int x)
{
      int l,r,mid;
      l=1,r=n+1;
      while (r-l>1)
      {
              mid=r+l>>1;
              if (P[mid].w>x) r=mid;
                         else l=mid;
      }
      if (P[l].w!=x) return 0;
      return l;
}
bool dfs(int x)
{
      if (!x) return false;
      if (ans[P[x].id]) return true;
      ans[P[x].id]=1;   
      if (find(a-P[x].w))
        if (!dfs(find(a-P[x].w))) return false;
      if (!dfs(find(b-P[x].w))) return false;
      return true;
}
bool FindAns()
{
      int i;   
      memset(ans,0,sizeof(ans)); 
      for (i=1;i<=n;i++)
      { 
              if (find(a-P[i].w)) continue;
              if (!dfs(i)) return false;
      }
      puts("YES");
      for (i=1;i<n;i++) printf("%d ",ans[i]);
      printf("%d\n",ans[n]);
      return true;
}
int main()
{       
      int i;   
      scanf("%d%d%d",&n,&a,&b);
      for (i=1;i<=n;i++) scanf("%d",&P[i].w),P[i].id=i;
      sort(P+1,P+n+1);
      if (!FindAns()) puts("NO");      
      return 0;
}