Codeforces Round #268 (Div. 2) D. Two Sets [stl

8161957	2014-10-10 06:12:37	njczy2010	D - Two Sets	GNU C++	Accepted	171 ms	7900 KB
8156137	2014-10-09 17:26:01	njczy2010	D - Two Sets	GNU C++	Wrong answer on test 9	30 ms	2700 KB
8156046	2014-10-09 17:20:58	njczy2010	D - Two Sets	GNU C++	Time limit exceeded on test 1	1000 ms	2700 KB
8155943	2014-10-09 17:16:09	njczy2010	D - Two Sets	GNU C++	Wrong answer on test 9	31 ms	2700 KB
8154660	2014-10-09 16:09:13	njczy2010	D - Two Sets	GNU C++	Wrong answer on test 6	15 ms	2700 KB

set真是太好用了，55555，要好好学stl

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.

题解转自：http://blog.csdn.net/u011353822/article/details/39449071

看到2Set我还真以为用2Set解，后来想想应该是2分匹配图，结果图左右两边分不出来，构不出图，弱爆了。。。唉。。早上起来又掉rating了

看了别人的解题报告，用的是直接暴力，能在a里处理的都放a集合，否则放入b集合，在b里开始遍历，如果b-x不在就去a里拿，如果a中也没有就输出NO，艾玛。。。。。

事实证明有时候想太多也不好

还能用并查集做，膜拜~~~ http://blog.csdn.net/budlele/article/details/39548063

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<set>
 10 #include<string>
 11 //#include<pair>
 12 
 13 #define N 100005
 14 #define M 1000005
 15 #define mod 1000000007
 16 //#define p 10000007
 17 #define mod2 100000000
 18 #define ll long long
 19 #define LL long long
 20 #define maxi(a,b) (a)>(b)? (a) : (b)
 21 #define mini(a,b) (a)<(b)? (a) : (b)
 22 
 23 using namespace std;
 24 
 25 int n,a,b;
 26 map<int,int>mp;
 27 set<int>f,g;
 28 vector<int>c;
 29 int x;
 30 int res[N];
 31 int flag;
 32 
 33 void ini()
 34 {
 35     memset(res,0,sizeof(res));
 36     flag=1;
 37     mp.clear();
 38     f.clear();
 39     g.clear();
 40     c.clear();
 41     int i;
 42     int y;
 43     for(i=1;i<=n;i++){
 44         scanf("%d",&x);
 45         mp[x]=i;
 46         f.insert(x);
 47     }
 48     for(set<int>::iterator it=f.begin();it!=f.end();it++){
 49         y=*it;
 50         if(f.find(a-y)==f.end()){
 51             c.push_back(y);
 52             //c.push_back(a-y);
 53         }
 54     }
 55 
 56     for(vector<int>::iterator it=c.begin();it!=c.end();it++){
 57         f.erase(*it);
 58         g.insert(*it);
 59     }
 60 }
 61 
 62 void solve()
 63 {
 64    while(g.empty()!=1){
 65         set<int>::iterator it=g.begin();
 66         int y=*it;
 67         if(g.find(b-y)!=g.end()){
 68             res[ mp[y] ]=1;
 69             res[ mp[b-y] ]=1;
 70             g.erase(y);g.erase(b-y);
 71         }
 72         else{
 73             if(f.find(b-y)!=f.end()){
 74                 res[ mp[y] ]=1;
 75                 res[ mp[b-y] ]=1;
 76                 g.erase(y);
 77                 f.erase(b-y);
 78                 if(f.find( a-(b-y) )!=f.end()){
 79                     g.insert(a-(b-y));
 80                     f.erase(a-(b-y));
 81                 }
 82             }
 83             else{
 84                 flag=0;return;
 85             }
 86         }
 87 
 88     }
 89 }
 90 
 91 void out()
 92 {
 93     if(flag==0){
 94         printf("NO
");
 95     }
 96     else{
 97         printf("YES
");
 98         printf("%d",res[1]);
 99         for(int i=2;i<=n;i++){
100             printf(" %d",res[i]);
101         }
102         printf("
");
103     }
104 }
105 
106 int main()
107 {
108    // freopen("data.in","r",stdin);
109     //freopen("data.out","w",stdout);
110    // scanf("%d",&T);
111    // for(int ccnt=1;ccnt<=T;ccnt++)
112    // while(T--)
113     while(scanf("%d%d%d",&n,&a,&b)!=EOF)
114     {
115         //if(n==0 && k==0 ) break;
116         //printf("Case %d: ",ccnt);
117         ini();
118         solve();
119         out();
120     }
121 
122     return 0;
123 }