UVA 1252 十五 Twenty Questions

Consider a closed world and a set of features that are defined for all the objects in the world. Each feature can be answered with ``yes" or ``no". Using those features, we can identify any object from the rest of the objects in the world. In other words, each object can be represented as a fixed-length sequence of booleans. Any object is different from other objects by at least one feature.

You would like to identify an object from others. For this purpose, you can ask a series of questions to someone who knows what the object is. Every question you can ask is about one of the features. He/she immediately answers each question with ``yes" or ``no" correctly. You can choose the next question after you get the answer to the previous question.

You kindly pay the answerer 100 yen as a tip for each question. Because you don‘t have surplus money, it is necessary to minimize the number of questions in the worst case. You don‘t know what is the correct answer, but fortunately know all the objects in the world. Therefore, you can plan an optimal strategy before you start questioning.

The problem you have to solve is: given a set of boolean-encoded objects, minimize the maximum number of questions by which every object in the set is identifiable.

Input 

The input is a sequence of multiple datasets. Each dataset begins with a line which consists of two integers, m and n: the number of features, and the number of objects, respectively. You can assume 0 < m11 and 0 < n128. It is followed by n lines, each of which corresponds to an object. Each line includes a binary string of length m which represent the value (``yes" or ``no") of features. There are no two identical objects.

The end of the input is indicated by a line containing two zeros. There are at most 100 datasets.

Output 

For each dataset, minimize the maximum number of questions by which every object is identifiable and output the result.

Sample Input 

8 1 
11010101 
11 4 
00111001100 
01001101011 
01010000011 
01100110001 
11 16 
01000101111 
01011000000 
01011111001 
01101101001 
01110010111 
01110100111 
10000001010 
10010001000 
10010110100 
10100010100 
10101010110 
10110100010 
11001010011 
11011001001 
11111000111 
11111011101 
11 12 
10000000000 
01000000000 
00100000000 
00010000000 
00001000000 
00000100000 
00000010000 
00000001000 
00000000100 
00000000010 
00000000001 
00000000000 
9 32 
001000000 
000100000 
000010000 
000001000 
000000100 
000000010 
000000001 
000000000 
011000000 
010100000 
010010000 
010001000 
010000100 
010000010 
010000001 
010000000 
101000000 
100100000 
100010000 
100001000 
100000100 
100000010 
100000001 
100000000 
111000000 
110100000 
110010000 
110001000 
110000100 
110000010 
110000001 
110000000 
0 0

Sample Output 

0 
2 
4 
11 
9

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 const int M=(1<<11)+10;
 6 const int inf=0x3f3f3f3f;
 7 int p[150],n,m;
 8 int dp[M][M];
 9 char a[150];
10 
11 int dfs(int s1,int s2)
12 {
13     int cnt=0,i,j;
14     if(dp[s1][s2]<inf)
15         return dp[s1][s2];
16     for(i=0;i<n;i++)
17     {
18         if((p[i] & s1)==s2)
19         {
20             cnt++;
21         }
22     }
23 
24     if(cnt<=1)
25     {
26         dp[s1][s2]=0;
27         return 0;
28     }
29 
30     for(i=0;i<m;i++)
31     {
32         if(s1 & (1<<i))
33             continue;
34         int nex=s1 | (1<<i);
35         dp[s1][s2]=min(dp[s1][s2],max(dfs(nex,s2),dfs(nex,s2^(1<<i)))+1);
36     }
37     return dp[s1][s2];
38 }
39 
40 int main()
41 {
42     int i,j,k;
43     while(scanf("%d %d",&m,&n)!=EOF)
44     {
45         if(n==0 && m==0)
46             break;
47         memset(p,0,sizeof(p));
48         memset(dp,inf,sizeof(dp));
49 
50         for(i=0;i<n;i++)
51         {
52             scanf("%s",a);
53             for(j=0;j<m;j++)
54             {
55                 if(a[j]==‘1‘)
56                     p[i]=p[i] | (1<<j);
57             }
58         }
59 
60         printf("%d\n",dfs(0,0));
61     }
62     return 0;
63 }

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