@vjudge
状压
可以这样设计状态
$dfs(s,a)$表示当前已经猜测的集合，目前还没有确定的数在$s$集合里面的共同特征为$a$，还要猜几次
枚举下一次猜的位为$k$
对于当前状态的答案是
$sum=max(dfs(s|2^k,a),dfs(s|2^k,a|2^k))+1$

对于每个$k$，取答案的最小值

至于边界，若满足状态$(s,a)$的数只有1个，则不用再猜；若还有2个，则再猜一次；若还有更多，则$dfs$

Code：

#include <bits/stdc++.h>
#define maxn 2110
using namespace std;
int cnt[maxn][maxn], dp[maxn][maxn], power[25], vis[maxn][maxn], n, m, kase;
char s[maxn];

int dfs(int s, int a){
    
    
	if (cnt[s][a] == 1) return 0;
	if (cnt[s][a] == 2) return 1;
	if (vis[s][a] == kase) return dp[s][a];
	vis[s][a] = kase;
	int ans = m;
	for (int i = 1; i <= m; ++i)
		if (!(s & power[i - 1])){
    
    
			int S = s | power[i - 1], A = a | power[i - 1];
			ans = min(ans, 1 + max(dfs(S, A), dfs(S, a)));
		}
	return dp[s][a] = ans;
}

int main(){
    
    
	power[0] = 1;
	for (int i = 1; i <= 20; ++i) power[i] = power[i - 1] << 1;
	while (1){
    
    
		scanf("%d%d", &m, &n);
		if (m == 0 && n == 0) break;
		++kase;
		for (int i = 0; i < power[m]; ++i)
			for (int j = 0; j < power[m]; ++j) cnt[i][j] = 0;
		for (int i = 1; i <= n; ++i){
    
    
			scanf("%s", s + 1);
			int x = 0;
			for (int j = 1; j <= m; ++j)
				x = (x << 1) | (s[j] == '1');
			for (int j = 0; j < power[m]; ++j) ++cnt[j][j & x];
		}
		printf("%d\n", dfs(0, 0));
	}
	return 0;
}