Codeforce 468B Two Sets_综合

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).

   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

在此讲出此题的贪心做法，不过我并不会证明，我只看到了题中有个条件说每个数字不一样，我就想会不会有贪心做法，就贪心交了一发，先排序数列，从最小的数字开始，看是否能加入数字大的集合，能加入，这个数字和max（a，b）-这个数字都要置false，然后在判断能否加入小的集合，不然就直接判NO。中间过程全用map完成。

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<string>
#include<cstring>
#include<algorithm>
#include<fstream>
#include<queue>
#include<map>
#include<stack> 
#include<vector>
#include<cmath>
#include<iomanip>
#define rep(i,n) for(i=1;i<=n;i++)
#define MM(a,t) memset(a,t,sizeof(a))
#define INF 1e9
typedef long long ll;
#define mod 1000000007
using namespace std;
int n,b[10],ii;
int a[100020];
int res[100020];
map<int,bool> g;
map<int,int> index;
bool ff;
int main()
{int i,j;while(scanf("%d%d%d",&n,&b[0],&b[1])!=EOF){ff=true;g.clear(); index.clear();if(b[0]>b[1]) ii=0;else          ii=1;rep(i,n){int tmp;scanf("%d",&tmp);a[i]=tmp;g[tmp]=true;index[tmp]=i;}sort(a+1,a+n+1);rep(i,n)if(g[a[i]]){int t1=a[i],t2=b[ii]-t1,t3=b[1-ii]-t1;if(g[t2]){res[index[t1]]=ii;res[index[t2]]=ii;g[t1]=false; g[t2]=false;	}	else if(g[t3]){res[index[t1]]=1-ii;res[index[t3]]=1-ii;g[t1]=false;g[t3]=false;}else{ff=false;break;}}if(ff){printf("YES\n");rep(i,n-1) printf("%d ",res[i]);printf("%d\n",res[n]);	}else printf("NO\n");}return 0;
}