Codeforces Round #438 (Div. 1 + Div. 2 combined) C. Qualification Rounds

C. Qualification Rounds

Problem Statement

    Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of n problems, and they want to select any non-empty subset of it as a problemset.
    k experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.
    Determine if Snark and Philip can make an interesting problemset!

Input

    The first line contains two integers n, k (1 ≤ n ≤ 105, 1 ≤ k ≤ 4) — the number of problems and the number of experienced teams.
    Each of the next n lines contains k integers, each equal to 0 or 1. The j-th number in the i-th line is 1 if j-th team knows i-th problem and 0 otherwise.

Output

    Print “YES” (quotes for clarity), if it is possible to make an interesting problemset, and “NO” otherwise.
    You can print each character either upper- or lowercase (“YeS” and “yes” are valid when the answer is “YES”).

Examples

Example 1
    Input
        5 3
        1 0 1
        1 1 0
        1 0 0
        1 0 0
        1 0 0
    Output
        NO
Example 2
    Input
        3 2
        1 0
        1 1
        0 1
    Output
        YES

Note

    In the first example you can’t make any interesting problemset, because the first team knows all problems.
    In the second example you can choose the first and the third problems.

题意

    给定n,k,表示有n道题和k支队伍。之后给出一张表格，第i行第j列若为1表示第j个队伍知道第i个题目，为0则表示不知道，问你是否能选出一套题（任意多），来满足任何一支队伍最多只知道其中一半的题目。

思路

    首先我们可以知道，所有能选出题目集合的方法中，都可以转化为只有两道题而所有队伍最多只知道一道，又因为k很小，只有4，所以我们可以$2^4*2^4$通过状压枚举一下就行了..

Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline void readInt(int &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
inline void readLong(ll &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
/*================Header Template==============*/
int cnt[17],x,t,n,k;
int main() {
    readInt(n);
    readInt(k);
    for(int i=1;i<=n;i++) {
        t=0;
        for(int j=1;j<=k;j++) {
            readInt(x);
            t=(t<<1)|x;
        }
        cnt[t]++;
    }
    for(int i=0;i<16;i++)
        for(int j=0;j<16;j++)
            if(cnt[i]&&cnt[j]&&(i&j)==0) {
                puts("YES");
                return 0;
            }
    puts("NO");
    return 0;
}