Codeforces Round #438 C
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思路:一定是选2个是最优的,将每一组化成二进制,题意即是选一些数出来,使得a1&a2&a3..&ak==0,显然,选的数字越少越好
AC代码:
#include "iostream" #include "iomanip" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #pragma comment(linker, "/STACK:102400000,102400000") #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a,x) memset(a,x,sizeof(a)) #define step(x) fixed<< setprecision(x)<< #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define ll long long #define endl ("\n") #define ft first #define sd second #define lrt (rt<<1) #define rrt (rt<<1|1) using namespace std; const ll mod=1e9+7; const ll INF = 1e18+1LL; const int inf = 1e9+1e8; const double PI=acos(-1.0); const int N=1e5+100; int f[20], bit[N][5], n, k; int bt1[10], bt2[10]; int fun(int t){ int ret=0; for(int i=k-1,bi=1; i>=0; --i){ ret+=bi*bit[t][i]; bi<<=1; } return ret; } int main(){ cin>>n>>k; for(int i=1; i<=n; ++i){ for(int j=0; j<k; ++j){ cin>>bit[i][j]; } f[fun(i)]=1; //cout<<fun(i)<<endl; } for(int i=0; i<=(1<<(k+1)-1); ++i){ if(!f[i]) continue; for(int j=0; j<=(1<<(k+1)-1); ++j){ if(f[j] && (j&i)==0){ cout<<"YES\n"; return 0; } } } cout<<"NO\n"; return 0; }
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