Codeforces Round #438 C. Qualification Rounds 题解

C. Qualification Rounds
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
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 any interesting problemset, because the first team knows all problems.

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

前三种暴力判，第四种手写

#include <bits/stdc++.h>#define digit (ch <  '0' || ch >  '9')using namespace std;template <class T> inline void read(T &x) {    int flag = 1; x = 0;    register char ch = getchar();    while( digit) { if(ch == '-')  flag = -1; ch = getchar(); }    while(!digit) { x = (x<<1)+(x<<3)+ch-'0'; ch = getchar(); }    x *= flag;}const long long maxn = 1e5+10;map <string, long long> r;long long a[maxn][10],c[maxn],d[10];long long ans1,ans2,n,k,maxx;int main() {    read(n); read(k);    for(register long long i = 1; i <= n; i++) {        for(register long long j = 1; j <= k; j++)            read(a[i][j]), c[i] += a[i][j], d[j] += a[i][j], maxx = max(maxx, d[j]);        ans1 += (!c[i]) ? 1 : 0, ans2 += (c[i] == 1) ? 1 : 0;    }    if(maxx == n) { printf("NO"); return 0; }    if(ans2 > 0 || ans1 > 0) { printf("YES"); return 0; }    if(k < 4) { printf("NO"); return 0; }    for(register long long i = 1; i <= n; i++) {        string s = "";        for(register long long j = 1; j <= k; j++) s += char('0'+a[i][j]);        r[s]++;    }         if(r["1100"] > 0 && r["0011"] > 0) printf("YES");    else if(r["1010"] > 0 && r["0101"] > 0) printf("YES");    else if(r["1001"] > 0 && r["0110"] > 0) printf("YES");    else printf("NO");    return 0;}