Codeforces Round #268 (Div. 1)-----B. Two Sets

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本文介绍了一个关于将一组整数划分成两个集合的问题，并提供了一种解决方案。问题要求每个元素x若属于集合A，则a-x也必须属于A；若属于集合B，则b-x也必须属于B。文章给出了输入输出格式及示例，并附带了解题思路和AC代码。

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).

Sample test(s)

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.

意解:这道题有很多的解法,我是用BFS做的,但是比赛因为一点小错误fail system text. 我把a - x 不存在的暂时确定为b

集合,存进队列,然后BFS,但是要注意一个问题,但当前的数配对的数是a集合里面的话,我们就要把a集合里面的那

个数取到b集合,如果没发取的,则输出"NO",详细看代码吧!

AC代码:

#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
using namespace std;

const int M = 1e5 + 10;
map<int,int>ms;
queue<int>ls;
int ma[M];

int main()
{
    int n,a,b;
    scanf("%d %d %d",&n,&a,&b);
    for(int i = 0; i < n; i++)
    {
        scanf("%d", ma + i);
        ms[ma[i]] = 1;
    }
    for(int i = 0; i < n; i++)
        if(!ms[a - ma[i]]) ls.push(ma[i]);

    while(!ls.empty())
    {
        int u = ls.front(),tp;
        ls.pop();
        if(ms[u] > 0 && ms[a - u] == 0 && ms[b - u] == 0)
        {
            puts("NO");
            return 0;
        }
        --ms[u]; //表示已被取到b集合去了;
        --ms[b - u];
        tp = a - b + u;
        if(ms[tp]) ls.push(tp);  //如果我所取的那个数是a集合里的,在bfs把另一个数取出来.
    }
    puts("YES");
    for(int i = 0; i < n; i++)
        if(ms[ma[i]] == 1) printf("0 ");
        else printf("1 ");
    puts("");
    return 0;
}