Codeforces Round #438 C

Qualification Rounds

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<#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; jj){
            cin>>bit[i][j];
        }
        f[fun(i)]=1; //cout<
    }
    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;
}

 

转载于:https://www.cnblogs.com/max88888888/p/7635466.html