Codeforces868c qualification rounds

Posted by yjjr's blog on February 6, 2018

标签:位运算

Snark and Philip are preparing theproblemset for the upcoming pre-qualification round for semi-quarter-finals.They have a bank ofn problems, and they want to select any non-emptysubset of it as a problemset.

kexperienced teams are participating in the contest. Some of these teams alreadyknow some of the problems. To make the contest interesting for them, each ofthe teams should know at most half of the selected problems.

Determine if Snark and Philip can make aninteresting problemset!

Input

The first line contains two integersn,k (1 ≤ n ≤ 105, 1 ≤ k ≤ 4) — the number of problems and the number of experiencedteams.

Each of the next n lines contains kintegers, each equal to 0 or 1. The j-th number in thei-th lineis 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

Input

5 3
1 0 1
1 1 0
1 0 0
1 0 0
1 0 0

Output

NO

Input

3 2
1 0
1 1
0 1

Output

YES

Note

In the first example you can't make anyinteresting problemset, because the first team knows all problems.

In the second example you can choose thefirst and the third problems.

题意:给定N行k列的矩阵,

求:能否从中挑选出若干行,使其组成的集合中不存在任何一列的和>=挑选的行数的一半

Step1首先当然想到的是贪心辣,显而易见,如果这个矩阵中存在一个肯定的答案,那么这个答案肯定可以为2行(如果全都是0的话也可以是一行)

可以按照每行的和从小到大排序,然后选定第一行,从前往后枚举第二行,如果存在一个正确得第二行,那么就输出yes

然而这种贪心的策略被卡了WA on test 129

这题竟然有那么多测试点,吃鲸

Step2 既然贪心不行,那就双重循环加判断来做呗,TLE on test 54……

心理崩溃qwq,这题数据那么毒瘤啊

Step3 前几天洪老师才讲的位运算唉,这题那么多0和1好像和位运算有些关系(这场CF洪老师也参加了,太劲辣%%%)

推式子,发现k<=4,这个范围很方便搞事情啊,于是二进制模拟,记录下每种01状态的数量,存在cnt数组内,然后如果这两种状态and=0,并且cnt数组都大于零(即存在这两种状态),那么就输出yes!

Code(正解)

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define mem(x,num) memset(x,num,sizeof x)
using namespace std;
int n,k,t,cnt[106],x;
int main()
{
	cin>>n>>k;
	rep(i,1,n){
		t=0;
	    rep(j,1,k){
	    	scanf("%d",&x);
	    	t=t*2+x;
	    }
	    cnt[t]++;
	}
	rep(i,0,(1<<k))
	    rep(j,0,(1<<k))
	        if(cnt[i]&&cnt[j]&&((i&j)==0)){cout<<"YES\n";return 0;}
	cout<<"NO\n";
	return 0;
}

code(模拟)
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define LL long long
#define mem(x,num) memset(x,num,sizeof x)
using namespace std;
const int maxn=100005;

struct node{int a,b,c,d,sum;}a[maxn];
int n,k,s[5];

inline int cmp(node x,node y){return x.sum<y.sum;}
inline bool check(int x,int y){
	if((a[y].a+a[x].a<=1)&&(a[y].b+a[x].b<=1)&&(a[y].c+a[x].c<=1)&&(a[y].d+a[x].d<=1))return true;
	else return false;
}

int main()
{
    scanf("%d%d",&n,&k);
	rep(i,1,n){
		if(k==1)scanf("%d",&a[i].a),a[i].sum=a[i].a,s[1]+=a[i].a;
		if(k==2)scanf("%d%d",&a[i].a,&a[i].b),a[i].sum=a[i].a+a[i].b,s[1]+=a[i].a,s[2]+=a[i].b;
		if(k==3)scanf("%d%d%d",&a[i].a,&a[i].b,&a[i].c),a[i].sum=a[i].a+a[i].b+a[i].c,s[1]+=a[i].a,s[2]+=a[i].b,s[3]+=a[i].c;
		if(k==4)scanf("%d%d%d%d",&a[i].a,&a[i].b,&a[i].c,&a[i].d),a[i].sum=a[i].a+a[i].b+a[i].c+a[i].d,s[1]+=a[i].a,s[2]+=a[i].b,s[3]+=a[i].c,s[4]+=a[i].d;
	}
	if(s[1]==n||s[2]==n||s[3]==n||s[4]==n){cout<<"NO\n";return 0;}
    sort(a+1,a+1+n,cmp);
    int temp=a[1].sum,j=1;
	while(a[j].sum==temp){ 
	    rep(i,1,n)
	        if(check(i,j)){cout<<"YES\n";return 0;}
	    j++;
    }
	cout<<"NO\n";
	return 0;
}




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