codeforces 468 B && 51nod 1557 两个集合

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本文介绍CF468B Two Sets问题的解决思路，使用map和排序进行有效分配，确保满足题目条件：若x属于集合A，则a-x也必须属于A；若x属于集合B，则b-x也必须属于B。提供了一种通过记录下标来优化解决方案的方法。

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).

   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.

题意：见http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1557

思路：map搞，开始想着遍历map两个map好像会出点问题？然后看的别人的用的记录下标的方法，要注意要优先考虑a、b中大的一方

#include<bits/stdc++.h>
using namespace std;

const int N = 1e5+5;
int p[N],myset[N];
map<int,int> ma;
map<int,int> id;
int main()
{
    int n,a,b;
    cin >> n >> a >> b;
    for(int i = 1;i <= n;i++)
        scanf("%d",&p[i]),ma[p[i]] = 1,id[p[i]] = i;
    sort(p+1,p+1+n);
    int gre = max(a,b),les = min(a,b);
    int ga,gb;
    if(gre == a)
        ga = 0,gb = 1;
    else
        ga = 1,gb = 0;
    for(int i = 1;i <= n;i++)
    {
        if(ma[p[i]])
        {
            if(ma[gre-p[i]])
                ma[p[i]] = ma[gre-p[i]] = 0,myset[id[p[i]]] = myset[id[gre-p[i]]] = ga;
            else if(ma[les-p[i]])
                ma[p[i]] = ma[les-p[i]] = 0,myset[id[p[i]]] = myset[id[les-p[i]]] = gb;
            else
                return 0 * printf("NO\n");
        }
    }

    printf("YES\n");
    for(int i = 1;i <= n;i++)
        printf("%d ",myset[i]);
    printf("\n");
    return 0;
}

在cf上面看到一些大佬的并查集写的 http://codeforces.com/contest/468/submission/7916826