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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.
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 < m 11and 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.
For each dataset, minimize the maximum number of questions by which every object is identifiable and output the result.
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
0
2
4
11
9
题意：这题意实在是很费解啊，每个串代表一个东西，这个东西是01组成的，现在你可以询问一个位置的特征，对于每个东西如果为1回答YES，如果为0回答NO。那么如果你当前无法区分开每个东西，就可以继续问一个特征，现在要求最坏情况下，你需要询问几次的最少次数。
思路：状态压缩DP+记忆化搜索，s1代表当前一共询问的特征的集合，s2答案的集合，计算当前s1&state[i] == s2的东西个数，由于要最坏情况，肯定是越大越好，当<=1的时候，说明已经全部区分开了，所以每次添加一个特征询问时候，回答可能是yes或no，要选其中大的，所以状态转移方程为：
dp[s1][s2] = min(dp[s1][s2], max(DP(s1|(1<<i), s2), DP(s1|(1<<i), s2^(1<<i))) + 1);
代码：
#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
const int M = 12;
const int N = 130;
const int MAXM = (1<<M);
int m, n, state[N], dp[MAXM][MAXM];
int DP(int s1, int s2) {
if (dp[s1][s2] != -1) return dp[s1][s2];
dp[s1][s2] = INF;
int num = 0, i;
for (i = 0; i < n; i++) {
if ((s1&state[i]) == s2)
num++;
}
if (num <= 1)
return dp[s1][s2] = 0;
for (i = 0; i < m; i++) {
if (s1&(1<<i)) continue;
dp[s1][s2] = min(dp[s1][s2], max(DP(s1|(1<<i), s2), DP(s1|(1<<i), s2^(1<<i))) + 1);
}
return dp[s1][s2];
}
int main() {
while (~scanf("%d%d", &m, &n) && n + m) {
char str[M];
for (int i = 0; i < n; i++) {
scanf("%s", str);
int s = 0;
for (int j = 0; str[j]; j++) {
if (str[j] == '1')
s = (s|(1<<j));
}
state[i] = s;
}
memset(dp, -1, sizeof(dp));
printf("%d\n", DP(0, 0));
}
return 0;
}
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