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 << "]" << "\n";
     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, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , 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-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e18 + ;
 const int maxN = 1e4 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 struct BitMap{
     char bm[maxN >> ];

     // 把第 x 位设置为 1
     void set(int x){
         bm[x >> ] |= 0x80 >> (x & 0x07);
     }
     // 把第 x 位设置为 1
     void clear(int x){
         bm[x >> ] &= ~(0x80 >> (x & 0x07));
     }
     // 获得第 x 位值
     bool get(int x){
         return bm[x >> ] & (0x80 >> (x & 0x07));
     }

     bool operator& (const BitMap &x) const{
         Rep(i, maxN >> ) if(bm[i] & x.bm[i]) return true;
         return false;
     }

     bool operator| (const BitMap &x) const{
         Rep(i, maxN >> ) if(bm[i] | x.bm[i]) return true;
         return false;
     }
 };

 int m, n, s;
 BitMap bitMask[];
 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 + , m - ) {
             if(bitMask[i] & bitMask[j]) continue;
             ans = false;
             i = m;
             break;
         }
     }

     if(ans) cout << "possible" << endl;
     else cout << "impossible" << endl;
     return ;
 }