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

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,0,sizeof(res));
    flag=1;
    mp.clear();
    f.clear();
    g.clear();
    c.clear();
    int i;
    int y;
    for(i=1;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()!=1){
        set<int>::iterator it=g.begin();
        int y=*it;
        if(g.find(b-y)!=g.end()){
            res[ mp[y] ]=1;
            res[ mp[b-y] ]=1;
            g.erase(y);g.erase(b-y);
        }
        else{
            if(f.find(b-y)!=f.end()){
                res[ mp[y] ]=1;
                res[ mp[b-y] ]=1;
                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=0;return;
            }
        }

    }
}

void out()
{
    if(flag==0){
        printf("NO
");
    }
    else{
        printf("YES
");
        printf("%d",res[1]);
        for(int i=2;i<=n;i++){
            printf(" %d",res[i]);
        }
        printf("
");
    }
}

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 0;
}