CodeForces 1166E The LCMs Must be Large

题目链接：http://codeforces.com/problemset/problem/1166/E

说明

1. LCM(一个集合) 为这个集合中所有元素的最小公倍数。
2. 如果$A subseteq B，LCM(A) leq LCM(B)$。 

题目大意

　　给定由 n 个整数组成的集合 A 。现给定 m 组集合，每个集合 S_i 都是 A 的一个真子集，求是否存在集合 A 使得对$forall_{1 leq i leq m} 不等式LCM(S_i) > LCM(A - S_i)恒成立$。

分析

　　考虑任意两个不同集合 S_i 和 S_j，它们有两种可能情况：

1. 无交集：$LCM(S_i) geq LCM(A - S_i) geq LCM(S_j) 和 LCM(S_j) geq LCM(A - S_j) geq LCM(S_i)矛盾$，所以只要有两个集合没有交集，A就不存在。
2. 有交集：有交集一不一定存在 A 呢？不晓得，只能说可能，反正题目只需要输出可不可能。

　　PS：用 set 做超时，自己写位图吧。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< string, int > PSI;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e18 + 7;
const int maxN = 1e4 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

struct BitMap{
    char bm[maxN >> 3];
    
    // 把第 x 位设置为 1 
    void set(int x){
        bm[x >> 3] |= 0x80 >> (x & 0x07);
    }
    // 把第 x 位设置为 1 
    void clear(int x){
        bm[x >> 3] &= ~(0x80 >> (x & 0x07));
    }
    // 获得第 x 位值 
    bool get(int x){
        return bm[x >> 3] & (0x80 >> (x & 0x07));
    }
    
    bool operator& (const BitMap &x) const{
        Rep(i, maxN >> 3) if(bm[i] & x.bm[i]) return true;
        return false;
    }
    
    bool operator| (const BitMap &x) const{
        Rep(i, maxN >> 3) if(bm[i] | x.bm[i]) return true;
        return false;
    }
};

int m, n, s;
BitMap bitMask[57];
bool ans = true;

int main(){
    INIT(); 
    cin >> m >> n;
    Rep(i, m) {
        cin >> s;
        Rep(j, s) {
            int x;
            cin >> x;
            bitMask[i].set(x);
        }
    } 
    
    Rep(i, m) {
        For(j, i + 1, m - 1) {
            if(bitMask[i] & bitMask[j]) continue;
            ans = false;
            i = m;
            break;
        }
    }
    
    if(ans) cout << "possible" << endl;
    else cout << "impossible" << endl;
    return 0;
}