CodeForces 1166E The LCMs Must be Large

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

说明

LCM(一个集合) 为这个集合中所有元素的最小公倍数。
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：

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。
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 ;

 }