Codeforce 868 C Qualification Rounds (计数原理）

最新推荐文章于 2019-03-09 00:15:53 发布

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最新推荐文章于 2019-03-09 00:15:53 发布

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分类专栏： ACM Problems

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本文针对CodeForces上的一个特定问题进行了解析，提出了一种寻找最少数量互不相交问题的方法。通过直接查找所有可能的问题组合，并利用位运算来判断两组问题是否相交，最终确定是否可以找到符合条件的问题集合。

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we need a lot of patience on this question
:http://codeforces.com/problemset/problem/868/C

Let’s talk about the method ,we can only find two question that has no intersection then that ‘ll be it!
because we want to choose minimum of question.If we can chose one question that no team knows it , we can definitely chose it . And if there are two more question that make a good set ,just two of them is fine.

if we directly fin all the question , we will get a tle then we
can iter all the situation of teams knows problem

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int cnt[20];
int main()
{
    ios::sync_with_stdio(0);
    int x,n,k;

    cin>>n>>k;
    for(int i = 0;i<n;i++)
    {
        int s = 0;
        for(int j = 0;j<k;j++)
        {
            int x = 0;
            cin>>x;
            s  = s * 2+x;
        }
        cnt[s]++;
    }
    for(int i = 0;i<(1<<k);i++)
    {
        for(int j = 0;j<(1<<k);j++)
        {
            if(cnt[i]>0&&cnt[j]>0 && (i&j)  == 0)
            {
                return cout<<"YES"<<endl,0;
            }
        }
    }
    cout<<"NO"<<endl;
    return 0;
}