Ultraviolet A 1252-Twenty Questions (memory-based search, State-compressed DP)

Link to the question: Ultraviolet A 1252
Question:
There are n binary strings with a length of m, each of which is different.
To separate all strings, you can ask whether a string is 0 or 1 at a time.
The minimum number of questions asked. All strings can be distinguished.
Thought Source: Click to open the link
Idea: m is very small. You can consider State compression. DP [S1] [s2] indicates the minimum number of steps required when the query status is S1. When the number of bits to be queried = S2 is smaller than 2, it can be distinguished. DP [S1] [s2] = 0; if it cannot be distinguished, ask again, s1 | = (1 <k). If you ask 0, S2 remains unchanged. If you ask 1, S2 | = (1 <K ). Recursively, you can find the answer.
Code:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#define maxn 1005#define MAXN 1000005#define mod 100000000#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int n,m,ans,tot,flag;char s[150][15];int dp[1<<11][1<<11],mp[150];int dfs(int s1,int s2){    if(dp[s1][s2]<INF) return dp[s1][s2];    int i,j,t=0;    for(i=1;i<=n;i++)    {        if((mp[i]&s1)==s2) t++;    }    if(t<=1)    {        dp[s1][s2]=0;        return 0;    }    int best=INF;    for(i=0;i<m;i++)    {        if(s1&(1<<i)) continue ;        best=min(best,max(dfs(s1|(1<<i),s2),dfs(s1|(1<<i),s2|(1<<i))));    }    dp[s1][s2]=best+1;    return best+1;}int main(){    int i,j,t;    while(~scanf("%d%d",&m,&n))    {        if(m==0&&n==0) break ;        for(i=1;i<=n;i++)        {            scanf("%s",s[i]);            mp[i]=0;            for(j=0;j<m;j++)            {                if(s[i][j]=='1') mp[i]|=(1<<j);            }        }        memset(dp,0x3f,sizeof(dp));        ans=dfs(0,0);        printf("%d\n",ans);    }    return 0;}