2019 ICPC Asia Taipei-Hsinchu Regional Problem J Automatic Control Machine (DFS,bitset)

• 题意:给你\(m\)个长度为\(n\)的二进制数,求最少选多少个使它们\(|\)运算后所有位置均为\(1\),如果不满足条件,则输出\(-1\).

• 题解:这题\(n\)的范围很大,所以我们先用\(string\)读,然后再转化为\(bitset\),之后再直接dfs爆搜,对于满足条件的维护一个最小值即可.

• 代码:

  #include <iostream>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <bitset>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  
  int t;
  int n,m;
  string s;
  int ans=INF;
  bitset<600> a[N];
  
  void dfs(int x,bitset<600> now,int cnt){
  	if(now.count()==n){
  		ans=min(ans,cnt);
  		return;
  	}
  	for(int i=x;i<m;++i){
  		bitset<600> tnow=now|a[i+1];
  		dfs(i+1,tnow,cnt+1);
  	}
  	return;
  }
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);
  	cin>>t;
  	 while(t--){
  	 	cin>>n>>m;
  	 	for(int i=1;i<=m;++i){
  	 		cin>>s;
  	 		a[i]=bitset<600>(s);
  	 	}
  	 	ans=INF;
  	 	bitset<600> now(0);
  	 	dfs(0,0,0);
  	 	if(ans==INF) ans=-1;
  	 	cout<<ans<<endl;
  	 }
  
      return 0;
  }