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!
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.
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").
5 3
1 0 1
1 1 0
1 0 0
1 0 0
1 0 0
NO
3 2
1 0
1 1
0 1
YES
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.
这题比赛时1a了， 当时还挺开心，觉得自己暴力写了个O(N)的代码。结果结束后看人家的ac代码，顿时感觉自己是个zz。
我在想思路的时候，已经用四位数（题目指出最多四个队伍，我们可以假设一直都有四个队伍，也是不影响的）来表示了，但是还是没想到用二进制。
最后只能一点一点的用if来判断。
附我的ac代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <string> 5 #include <algorithm> 6 #include <cmath> 7 using namespace std; 8 const int maxn = 100005; 9 int nu[maxn][6]; 10 int cnt[22]; 11 int main() { 12 ios::sync_with_stdio(false); 13 int n,k; 14 cin>>n>>k; 15 for(int i = 0; i < n; i++) { 16 for(int j = 0; j < k; j++) { 17 cin>>nu[i][j]; 18 } 19 } 20 for(int i = 0; i < n; i++) { 21 if(nu[i][0] == 0 ) { 22 if(nu[i][1] == 0) { 23 if(nu[i][2] == 0) { 24 if(nu[i][3] == 0) { 25 cnt[16]++; 26 } 27 else { 28 cnt[1]++; 29 } 30 } 31 else { 32 if(nu[i][3] == 0) { 33 cnt[2]++; 34 } 35 else{ 36 cnt[10]++; 37 } 38 } 39 } 40 else { 41 if(nu[i][2] == 0) { 42 if(nu[i][3] == 0) { 43 cnt[3]++; 44 } 45 else { 46 cnt[9]++; 47 } 48 } 49 else { 50 if(nu[i][3] == 0) { 51 cnt[8]++; 52 } 53 else{ 54 cnt[14]++; 55 } 56 } 57 } 58 } 59 else { 60 if(nu[i][1] == 0) { 61 if(nu[i][2] == 0) { 62 if(nu[i][3] == 0) { 63 cnt[4]++; 64 } 65 else { 66 cnt[7]++; 67 } 68 } 69 else { 70 if(nu[i][3] == 0) { 71 cnt[6]++; 72 } 73 else{ 74 cnt[13]++; 75 } 76 } 77 } 78 else { 79 if(nu[i][2] == 0) { 80 if(nu[i][3] == 0) { 81 cnt[5]++; 82 } 83 else { 84 cnt[12]++; 85 } 86 } 87 else { 88 if(nu[i][3] == 0) { 89 cnt[11]++; 90 } 91 else{ 92 cnt[15]++; 93 } 94 } 95 } 96 } 97 } 98 if(cnt[16]) cout<<"YES"<<endl; 99 else if(cnt[1]) { 100 if(cnt[2]||cnt[3]||cnt[4]||cnt[5]||cnt[6]||cnt[8]||cnt[11]) 101 cout<<"YES"<<endl; 102 else cout<<"NO"<<endl; 103 } 104 else if(cnt[2]) { 105 if(cnt[3]||cnt[4]||cnt[5]||cnt[7]||cnt[9]||cnt[12]) 106 cout<<"YES"<<endl; 107 else cout<<"NO"<<endl; 108 } 109 else if(cnt[3]) { 110 if(cnt[4]||cnt[6]||cnt[7]||cnt[10]||cnt[13]) 111 cout<<"YES"<<endl; 112 else cout<<"NO"<<endl; 113 } 114 else if(cnt[4]) { 115 if(cnt[8]||cnt[9]||cnt[10]||cnt[14]) 116 cout<<"YES"<<endl; 117 else cout<<"NO"<<endl; 118 } 119 else if(cnt[5]) { 120 if(cnt[10]) 121 cout<<"YES"<<endl; 122 else cout<<"NO"<<endl; 123 } 124 else if(cnt[6]) { 125 if(cnt[9]) 126 cout<<"YES"<<endl; 127 else cout<<"NO"<<endl; 128 } 129 else if(cnt[7]) { 130 if(cnt[8]) 131 cout<<"YES"<<endl; 132 else cout<<"NO"<<endl; 133 } 134 else cout<<"NO"<<endl; 135 136 return 0; 137 }
附别人的ac代码：
1 #include<cstdio> 2 int n,k,a[17],b[5]; 3 int main(){ 4 scanf("%d%d",&n,&k); 5 while(n--){ 6 for(int i=1;i<=k;i++)scanf("%d",&b[i]); 7 a[b[1]+b[2]*2+b[3]*4+b[4]*8]=1; //化为二进制 8 } 9 for(int i=0;i<=(1<<k);i++) 10 for(int j=i;j<=(1<<k);j++) 11 if(!(i&j)&&a[i]&&a[j]) //如果存在i和j四位都不同时为1 例：1010 0100 12 {puts("YES");return 0;} 13 puts("NO"); 14 return 0; 15 }