Codeforces Round #268 (Div. 2) D. Two Sets [stl - set + 暴力]

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

D. Two Sets

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

 

 

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

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<set>
 #include<string>
 //#include<pair>
 
 #define N 100005
 #define M 1000005
 #define mod 1000000007
 //#define p 10000007
 #define mod2 100000000
 #define ll long long
 #define LL long long
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)
 
 using namespace std;
 
 int n,a,b;
 map<int,int>mp;
 set<int>f,g;
 vector<int>c;
 int x;
 int res[N];
 int flag;
 
 void ini()
 {
     memset(res,,sizeof(res));
     flag=;
     mp.clear();
     f.clear();
     g.clear();
     c.clear();
     int i;
     int y;
     for(i=;i<=n;i++){
         scanf("%d",&x);
         mp[x]=i;
         f.insert(x);
     }
     for(set<int>::iterator it=f.begin();it!=f.end();it++){
         y=*it;
         if(f.find(a-y)==f.end()){
             c.push_back(y);
             //c.push_back(a-y);
         }
     }
 
     for(vector<int>::iterator it=c.begin();it!=c.end();it++){
         f.erase(*it);
         g.insert(*it);
     }
 }
 
 void solve()
 {
    while(g.empty()!=){
         set<int>::iterator it=g.begin();
         int y=*it;
         if(g.find(b-y)!=g.end()){
             res[ mp[y] ]=;
             res[ mp[b-y] ]=;
             g.erase(y);g.erase(b-y);
         }
         else{
             if(f.find(b-y)!=f.end()){
                 res[ mp[y] ]=;
                 res[ mp[b-y] ]=;
                 g.erase(y);
                 f.erase(b-y);
                 if(f.find( a-(b-y) )!=f.end()){
                     g.insert(a-(b-y));
                     f.erase(a-(b-y));
                 }
             }
             else{
                 flag=;return;
             }
         }
 
     }
 }
 
 void out()
 {
     if(flag==){
         printf("NO\n");
     }
     else{
         printf("YES\n");
         printf("%d",res[]);
         for(int i=;i<=n;i++){
             printf(" %d",res[i]);
         }
         printf("\n");
     }
 }
 
 int main()
 {
    // freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
    // scanf("%d",&T);
    // for(int ccnt=1;ccnt<=T;ccnt++)
    // while(T--)
     while(scanf("%d%d%d",&n,&a,&b)!=EOF)
     {
         //if(n==0 && k==0 ) break;
         //printf("Case %d: ",ccnt);
         ini();
         solve();
         out();
     }
 
     return ;
 }